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User:Eml5526.s11.team7.jb/work57p2

Solution
The Lagrangian Element Basis Function (LEBF) is given by the following equation and generates a polynomial of degree $$\displaystyle n-1$$:

 * $$L_{i,n}\left (x\right )=\prod_{k=1}^{n}\frac{x-x_{k}}{x_{i}-x_{k}},k\neq i$$

For a linear function the number of nodes is 2 ($$\displaystyle n=2$$). For uniform discretization with $$\displaystyle nel=4$$ the domain $$\displaystyle \Omega$$ will be subdivided into 4 equal length elements each having two nodes. The LEBFs for the first element $$\displaystyle n=1$$ over the elemental domain $$\displaystyle \omega^{(1)}=[0,.25]$$ can be written as follows:

 * $$L_{1,2}^{(1)}\left (x\right)=b_{1}^{(1)}=\frac{(x-x_2)}{x_1-x_2}=\frac{(x-.25)}{0-.25}=-4(x-.25)$$

 * $$L_{2,2}^{(1)}\left (x\right)=b_{2}^{(1)}=\frac{(x-x_1)}{x_2-x_1}=\frac{(x-0)}{.25-0}=4(x)$$

Recall the constraint breaking solution (CBS) states the basis functions must satisfy $$\displaystyle \left\{ b_i(x), i=0,1,2,...\right\}$$ such that $$\displaystyle b_0(\beta) \neq 0$$ and $$\displaystyle b_i(\beta) = 0$$ for $$\displaystyle i=1,2,...,n$$.

It can be seen that these basis functions for the first element satisfy the CBS at each node. For the first node, the basis function has a value of one at that node and a value of zero at every other node (there is only one other node for the linear LEBF). In other words the basis function is equal to the kronicker delta at each node.

 * $$b_{1}^{(1)}(\beta^{(1)}_1)=L_{1,2}^{(1)}\left (0\right)=b_{1}^{(1)}(0)=-4(0-.25)=1$$

 * $$L_{1,2}^{(1)}\left (.25\right)=b_{1}^{(1)}(.25)=-4(.25-.25)=0$$

Similarly for the second node, the basis function has a value of one at that second node and a value of zero at the first node.

A Matlab script was written to plot the LEBFs at each node for each element. The Matlab script is given followed by the plots. The first plot shows each of the LEBFs plotted across the entire domain. The second plot is more informative as the plot is divided into the elemental domains and lines corresponding to the values zero and one are present. The legend states the node number, $$n$$, and element number, $$e$$, of each LEBF.

As with the CBS, the coefficient at the essential boundary condition is equal to the essential boundary condition because the basis function is equal to one at the essential boundary condition:

 * $$g = \mathbf{d_{1}^{(1)}}\mathbf{b_{1}^{(1)}}(0)=\mathbf{d_{1}^{(1)}}*1=\mathbf{d_{1}^{(1)}}=4$$

The element matrices are formed as in problems 5.1 and 5.2 as follows:

 * $$\displaystyle \mathbf{F}^{e}_{F}=b_{i}^{(1)}(\alpha_{i}^{e} )h + \int_{\omega^{e}}^{.} {{b_{i}^{(1)}}fdx}$$

\displaystyle

h = {a_2}(\alpha )\frac = (2+3\alpha)\frac = 5*12/5=12

$$

 * $$\displaystyle \mathbf{F_{F}^{(1)}}=\mathbf{F}^{(1)}_{2}=b_{2}^{(1)}(.25)h + \int_{0}^{.25} {{b_{2}^{(1)}}fdx}=1*12+\int_{0}^{.25} 4x(5x)=12.104167$$

$$ \displaystyle

\therefore {F_F} = \left[ {\begin{array}{ccccccccccccccc} {12.104167} \end{array}} \right]

$$

 * $$\displaystyle \mathbf{K}^{e}_{ij}=\int_{\omega^{e}}^{.} b_{i}^{e'}(x^e)a_2(x^e)b_{j}^{e'}(x^e)dx$$

\displaystyle

\mathbf{K^{(1)}_{FE}} = \left\{ {\mathbf{K^{(1)}_{1j}};j = 2,...,n} \right\}

$$

\displaystyle

\mathbf{K^{(1)}_{12}}= \int\limits_0^{.25} b_{1}^{(1)'}a_2b_{2}^{(1)'}=\int\limits_0^{.25}-4*(2+3x)*4=-32(.25)-24(.25)^2=-9.5

$$

\displaystyle \therefore \mathbf{K^{(1)}_{FF}} = \begin{bmatrix} -9.5 \end{bmatrix} $$

\displaystyle

\mathbf{K^{(1)}_{FF}} = \left\{ {\mathbf{K^{(1)}_{ij}};i,j = 2,...,n} \right\}

$$

\displaystyle

\mathbf{K^{(1)}_{22}}= \int\limits_0^{.25} b_{2}^{(1)'}a_2b_{2}^{(1)'}=\int\limits_0^{.25}4*(2+3x)*4=32(.25)+24(.25)^2=9.5

$$

\displaystyle \therefore \mathbf{K^{(1)}_{FF}} = \begin{bmatrix} 9.5 \end{bmatrix} $$

 * $$\mathbf{K^{(1)}} \mathbf{d^{(1)}} = \mathbf{F_{F}^{(1)}} - \mathbf{K_{FE}^{(1)}} g$$

 * $$\mathbf{d_{2}^{(1)}} = \mathbf{K}_{FF}^{(1)-1}( \mathbf{F_{F}^{(1)}} - \mathbf{K_{FE}^{(1)}} g)$$

 * $$\therefore \mathbf{d_{2}^{(1)}} = 5.5208$$

Now the process is repeated for each element. The matrices are assembled into global matrices and all of the coefficients can be determined. The approximate solution is formed for each element by taking the coefficients times the basis functions. The approximate solutions for each element make up the global approximate solution.

 * $$\displaystyle \mathbf{K}_{ij} = \sum_{e=1}^{nel}\mathbf{K}^{e}_{ij}$$

 * $$\displaystyle \mathbf{F}_{i} = \sum_{e=1}^{nel}\mathbf{F}^{e}_{i}$$

 * $$\displaystyle u^{he}(x)= \sum_{j=1}^{n}d_jb_j(x),\quad \quad u^{h}(x)=\sum_{j=1}^{nel}u^{he}(x) $$

A Matlab script was written to generate the plots. The results and script are given below:

title: User:Eml5526.s11.team7.jb/work57p2


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


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User:Egm6321.f10.team2.oztekin/Newwwwww

= Problem 5.5 : Calculix-Using CCX module=

Disk Problem
After building disc in cgx module then we might want to know that properties of nodes,lines, surfaces, and bodies so now lets look at how we can get more information what we have done for our model.

First we have to open our program. To do this run the following ;

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

$$  \displaystyle [All Programs]\Rightarrow [bConverged]\Rightarrow [Calculix]\Rightarrow [Calculix Command] $$
 * style="width:95%" |
 * style="width:95%" |

 * }

Open calculix and go to directory where the basic examples stays in your harddrive.
 * {| style="width:100%" border="0"

$$  \displaystyle ...\Rightarrow [Program Files]\Rightarrow [bConverged] \Rightarrow [Calculix] \Rightarrow [Cgx] \Rightarrow [Examples] \Rightarrow [Basic] $$
 * style="width:95%" |
 * style="width:95%" |

 * }

Then type to open disc example:

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

$$  \displaystyle [cgx]\Rightarrow [disc] $$
 * style="width:95%" |
 * style="width:95%" |

 * }

Now we want to learn about point properties. To see point properties in our graph secren type

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

$$  \displaystyle \begin{matrix} plot & pa   & all \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |

 * }

We need to mesh disc to see the mesh properties to mesh the disc please type following steps

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

$$  \displaystyle \begin{matrix} elty & all & qu8 \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |

 * }

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

$$  \displaystyle \begin{matrix} mesh & all \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |

 * }

What we have done so far is that we opened disc and meshed it with quadratic elements. So we are ready to see element properties on our graph screen. To see properties of elements type,
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{matrix} plot & ma  & all \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |

 * }

Also we can get more specific properties form the special editor of calculix called 'SciTE.' We can go to examples directory and just basically right click on disc example and open with by editing. SciTE automatically will be opened.

From below we can easly see the properties of all components of disc. Especially point properties are boxed.

Generating Different Meshes
We will do the same procedures as we have done before. Our model is prepared. So we will mesh our model and take properties of different kinds of meshes for 2-D. Our meshes for 2-D are:

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

$$  \displaystyle \begin{matrix} tr3 & 3 node & 2D element \\ gu4 & 4 node & 2D element\\ qu8 & 8 node & 2D element \end{matrix} $$     (N)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{matrix} Mesh & with  & tr3 \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |

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

$$  \displaystyle \begin{matrix} Mesh & with  & qu4 \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |

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

$$  \displaystyle \begin{matrix} Mesh & with  & qu8 \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |

 * }

CCX module
In this module we will analyze for unknowns that we are interested in. This unknowns can be displacements or temperatures. Also we can find fluxes between these nodes.

For the first step go to directory where beam example stays in your harddrive.

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

$$  \displaystyle ....[Program Files]\Rightarrow [bConverged]\Rightarrow [Calculix]\Rightarrow [ccx]\Rightarrow [test] $$
 * style="width:95%" |
 * style="width:95%" |

 * }

Find beampl.inp and right click on it. Then edit it by selecting open with. SciTe automatically will be opened. Every command inside is ready so we have assumed the we have completed by the mesh and ready for solving.So click tools and click 'solve'.

You have to see that $$ \displaystyle Job Finished $$  Then click tools and click 'post prosess'.

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

$$  \displaystyle Result 1 $$
 * style="width:95%" |
 * style="width:95%" |

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

$$  \displaystyle Result2 $$
 * style="width:95%" |
 * style="width:95%" |

 * }

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Complex socio-ecological systems/Conference discussion

March 25th

Conference Discussion and Discussion of Scheffer et al 2000

Facilitators: Bob, Sam, Simone and Flavia

Required reading:

Scheffer, M, Willliam Brock and Frances Westley. 2000 Socioeconomic Mechanisms Preventing Optimum Use of Ecosystem services. Ecosystems 3: 451-471.

Long summary of Scheffer et al.'s paper:

This article has two parts: 1) A discussion of how ecosystems work, and 2) A discussion of the social elements surrounding ecosystems. Within the discussion of the social elements, the authors point out how stakeholder behavior depends on the relationship of the stakeholder to the ecosystem (affectors versus enjoyers), on the interaction among people within a group of stakeholders (collective action and free riding), and on the level of social organization of each group of stakeholders. As a result of the interplay of natural and social, societies tend to move towards suboptimal compromises that are not good for the ecosystem and to the welfare of society.

1) As regards the functioning of ecosystems, the authors explain that ecosystems are complex, having multiple stable states and nonlinear responses to human use. Because of nonlinearity, changes to ecosystems are often not smooth. When ecosystems change, the change occurs quickly and is often irreversible. Authors point out that ecosystem change can often be examined by the study of key state variables. The authors refer to the human use of ecosystem as the "stress" caused by humans. They explain that point the nonlinear ecosystem response to stress can lead to unexpected catastrophic shifts. They authors also explain the implications of catastrophic shifts. To illustrate their ideas, the authors discuss what studies of shallow lakes have taught us about ecosystems. One important conclusion from these studies is that lakes have two alternative stable states that depend on the nutrient levels, water turbidity and amount of plants in the lakes. The two alternative stable states are represented as the bottom of two valleys in the authors' graphs. Authors explain that once lakes reach a critical level of turbidity and high levels of nutrient, trying to reverse the lake to previous levels of nutrient and turbidity is not simple since "the system tends to say tin the "turbid valley of attraction" (p. 456).

2) Authors discuss the social elements surrounding lake ecosystems. They present possible stakeholders who have a "stake" in the conditions of the lake (e.g. fishermen, farmers, households, water companies). The authors introduce two concepts:"welfare" or "utility," which are used somewhat as synonyms. The authors argue that estimating the welfare function for each stakeholder is not simple. Authors also question whether society should focus on maximizing the value of a lake for human use, and ignore other concerns, such as ethical concerns. Author explains what a monopolistic owner of a lake would do - the owner would seek to maximize the sum of values that he or she can extract from the lake. The owner would also have an incentive to preserve the lake in order to keep profiting from the lake. Authors argue that is important to avoid cost-benefit analysis and "focus the discussion on how society might measure and extract all potential values out of the bundle of resources comprised by the lake and its watershed" (p. 457).

The authors explain that there are different services that can be extracted from a lake and that there is a trade-off between different lake uses. The authors talk about a hypothetical RASP (Rational Social Planner). They argue that a RASP should know the value of a system for different users in order to find the integrated use that brings the most welfare for society. The RASP would be able to do this by combining the ecosystem response curves with the welfare functions of the lake. The authors distinguish two types of stakeholders: the "affectors", who have a large direct impact on the conditions of an ecosystem, and the "enjoyers," who benefit from an ecosystem, but do not directly mess up the ecosystem through their activities. Authors recognize that in practice there are many types of stakeholders, who often have overlapping rather than complementary interests. The authors combine the maximum utility graph with the stable equilibrium lines to explain that that compromise solutions between the affectors and enjoyers are not good for the stakeholders or for the ecosystem. Author distinguishes overall welfare from the optimum utility solutions that result from compromises and political pressure. The authors also distinguish the "good" and "bad" side of the utility curve, which correspond respectively to good ecosystem conditions and bad ecosystem conditions.

Therefore, RASPs should not seek an optimum solution that reflect a compromise between affectors and enjoyers. The authors explain that the political process weighs in favor of affectors, who are usually more organized as a group and can therefore pressure decision makers more effectively. As results, political decisions regarding ecosystems tend to benefit more affectors in detriment of enjoyers and the ecosystems. The authors therefore note that political pressure is a strong determinant of what decisions are actually make as regards ecosystem use.

Authors use the lake example to explain theories about what social elements bring societies closer or father from obtaining optimum utility from an ecosystem. Regarding the social elements, the authors recognize that reality is more complex than the simplifications they make. Still, they talk about a lot of imaginary situations based on "ideal" situations and "ideal" behaviors. The authors argue that to make smarter decisions regarding ecosystem use, authorities should have insights on how ecosystems work and adjust policies according to ecosystem and societal variations. Still, the interactional elements of society lead decision makers away from passing regulations to guide resources users towards sustainable use. The authors provide the example of taxation, one method that could be used to regulate ecosystem use. In this case, the RASP would establish a fixed tax rate per unit of pollution. The result of the tax is that after a certain level of pollution, with diminishing returns to investments, it would be too costly for affectors to continue polluting. This would lead affectors to slow down pollution, which would be good for the ecosystem and for the enjoyers. However, one problem is that affectors have greater organizational capacity than enjoyers and they can mobilize to stop the RASP from approving the tax. The authors use game theory and theory on collective action to explain power differentials across stakeholders and the different ability of stakeholders to overcome collective action problems (free rider problem). Specifically, the authors explain why affectors can exert greater political pressure on decision makers than enjoyers, which is a result of the imbalance of costs and benefits between affectors and enjoyers, and their unequal level of social organization (social capital). Finally, the authors identify a problem that makes it harder for a well-intended RASP to plan towards optimum ecosystem: Slow social dynamics, that is, stakeholders take a while to notice a significant ecosystem variation, to negotiate and to respond to it. Authors suggest that greater social networking among stakeholders and a common culture of reciprocity can speed up the social dynamics and result in more prompt responses to ecosystem changes.


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Talk:Complex socio-ecological systems/Conference discussion

This week we have 2 discussion threads, the Resilience 2011 conference and on the paper by Scheffer et al. Later, we will place these on separate pages.

Resilience 2011 conference

Very evident that "Resilience / Panarchy" is a young, developing discipline. The conference was an opportunity to drink deeply from that cup, and I feel like I got a good sense of the field. For me, the key scientific innovation of resilience is that it calls attention to the surprise, non-linear, "tipping points" in nature. Manage for surprise, expect the unexpected. This is a paradigm shift in terms of how we look at nature (goes beyond the equilibrium, "balance of nature" paradigm).

The other paradigm shift, which I see as a noble but insufficient effort, seeks to integrate natural science and social science. It's clearly natural scientists trying to bring about this integration, so naturally they take natural science concepts and apply them to social systems. This effort is incomplete and at some levels naive. There was a certain reaction at the conference from the social science side ... 'you guys are naive ... your concepts will not apply in social systems.' Specifically, the social scientists resent having ecological concepts applied to social systems because they do not take account of agency, reflexivity, and power differentials. It was good to see this debate, though probably any social scientist participating in the conference would say that it was under-represented.

My personal evaluation of this effort by the natural scientists to integrate social systems into their analysis: 1) it is essential 2) although insufficient to date, it is possible, from an analytical perspective -- they are making strides, that need to be extended 3) in contrast to their social analysis, the resilience community's process of interacting with stakeholders is even less developed. bordering on the negligent. The "Resilience Assessment" tool, which we are seeking to apply in our Cotrigauçu work, is very expert-driven, a typical technocratic, planning approach.  I think our group in Mato Grosso is innovating in this regard, and has much to contribute to the Resilience network.  We got a wonderful response to our presentation.  Our project is now registered at the Resilience Alliance web site of assessment projects (last item on this page http://www.resalliance.org/cdirs/raprojects/index.php/0/browse)

Shifting paradigms, shifting grounds, integrating to transform, innovating to survive. The conference brought some very clear answer to our questions, and at the same time, did not relieve our anxieties. It confirmed that the social sciences are under-represented and that the resilience approach has a clear focus on maintaining ecosystems functioning and services. But how do you work with a multitude of perspectives and interests of actors involved in the management of ecosystems is still very unclear, and we can't start to develop a new science or paradigm for this, because there is already a long road driven by other knowledge fields, such as anthropology, development theories and participation. Scheffer's brilliant presentation on "Holism 2.0" calls for defragmentation of sicence. Here is where complexity has a role to play: the whole is bigger than the sum of parts, but the parts "do" exist" and should be considered as such. The good news is the attempt to bridge different disciplines, views and "people", who are the ones capable of transforming things in different scales, local to global and back if you wish. No reinventing the square whell needed. Integration needed. Resilience is no panacea, and the bigger theory is systems thinking. Learning with history, coping with the unexpected through building stronger societies, to prepare for change. Clear need for working in urban areas, bringing art, emotions and science into play. It was very interesting for me the possibility of meeting the authors and asking questions directly toi them and verifying that they do have sifferent opinios, bu some common though, for instance, talking about interdisciplinarity as learning with local actors, practitioners and between academics with a will to do so. Another field of relevance in my work is, if we are to apply the concept in social systems, it makes sense for me to think about cultural resilience as a useful concept for indigenous societies in a changing world. I believe that is can be a concept that enables continuity and change, and in which longitudinal studies allow for understanding the path, mechanisms and outcomes of being aware of the changing or transformative process. It can't and won't replace studies of dynamics of cultural identity in anthropology. But is might be an interesting complementary concept to explore and build along with other disciplinary fields, promoting learning between disciplines, scientists, indigenous and other stakeholders. Simone Brazil nut.

- I have a multitude of impressions to share about this conference, but I would like to use an extended metaphor, maybe more of an allegory, as indeed that is what the ResAlliance is in the business of doing.

Let's say a small band of clerics, we'll call them The Chosen (the word illuminati was, well, chosen already), decide to engage on a noble pursuit: a new dogma that helps to explain the workings of the galaxy. They call it the Sign of the Infinity (or Horizontal Eight depending on which gospel you select from), and this dogma can pretty much describe why you prefer blue M&Ms, where baby rabbits go when they die, and how radioactive that plant in Fukushima really is. Really, it can do all that. The Chosen are effective proselytizers: they are positioned in high places and have earned themselves good standing in their local parishes. They are very effective with the Partly Elect, because they have a similar origin myth, but they are less effective with the Heathens, who believe in strange ideas about spiritual representation and the ghastly god called On-Toh-Logee. Until one day they are able to make headway with a particularly powerful band of the Heathens, who they buy off with promises of whiskey and ammunition, and they begin to plot the realization of their full vision: a world where the Sign is posted on every rooftop and staked in every yard. One day, in the middle of a highly charged seance, The Chosen begin to realize that not all of the heathens have been fully converted. For instance one heathen is observed carrying the false idol On-Toh-Logee in his hand; another is heard uttering occult words that no civilized ears should deign to hear. But the seance continues, and little is made of the disruptions by the high priests. When the event is over, and everyone has arisen from their fevered, catatonic state, The Chosen are more emboldened and feel that their mission must be accomplished faster and more efficiently. But some heathens are not convinced. They meet quietly outside the revival tent and begin to ask questions: "Should we throw away the whiskey?" asks one. "Should we use the ammunition (if only we had guns!)?" inquires another.

I don't think I'm as cynical as I portray here, but I thought I should depict at least one reading of the conference events, o best beloved. ~ Sam

- The conference calls us to the reality of the academic world, with remarkable exceptions. First science is driven by financial resources, second publishing is the most important outcome, third high resilience of disciplines, and good scientific material could be lost in the mismatch use of concepts. We saw all of it in one place. However the good superposed the bad in being there. First going as a group was a excellent source of maturing what we learned along the day. We had excellent discussion about all the problems I presented above, and how we as group could make better, avoiding this traps. Second was great see in face some of the persons that wrote most of the theory. Make all the difference when I read a paper wrote by Orston, have her image associated. I don’t know why, but make difference. Third because was a huge conference, we were able to know the state of art of resilience, what are the constraints, fragilities, advances and groups that are leading the process. About the academic knowledge, for me the most remarkable was the increase use of SNA to study human systems, the difficulty that is work together social and natural sciences, the fragility that represents the use of resilience science to answer stakeholder’s needs. Ricmel 12:53, 25 March 2011 (UTC)

paper by Scheffer et al

Scheffer may have been the most stimulating speaker at the Resilience conference. Right Sam? I found figures 1 and 2 to be a good review ... very didactic. The meat of this paper is the second half, that gives a very rigorous analysis of interest group politics, and why management will always aim for an optimum, overshoot the optimum, and why this can often lead to collapse. - The Scheffer et al. piece reads like the opening salvo to the siege that PES has taken on modern environmental science and policy. The premise is simple: while the biophysical system degrades, the social system continues on a growth trajectory. Or as Carpenter et al. 2001 put it way back when we were reading them: "The collapse of the biophysical system does not usually cause a simultaneous collapse of the local economy because the economy does not solely depend on the lake. The agricultural sector, for example, may use the lake only as a sink for wastes while extracting the water needed for agriculture from the groundwater.Other interest groups may likewise depend wholly or in part on resources from other regions."

Indeed, this piece points to a lot of the material that is taken as gospel in SES-type research these days, with a particular focus on those northern European lakes we love so dearly. They include the slow dynamics of soil and sediment phosphorous, exclusion of water quality from market valuation, the value of the ecosystem services not being properly accounted in indices of social wellbeing, a lack of a market for pollutant discharge (cap and trade), the fact that farmers do not have incentives to control pollution, and the mis-matches of political power between environmental advocates and business interests.

Read from the vantage point of 2000, it appears fresh, but read from the vantage point of today it almost seems as though the adaptive cycle provides nothing new here. Certainly there are good frames to explore PES, and the social and environmental conditions which constitute, without using resilience frameworks? Scheffer's talk was much along the same lines, and Bob pointed out that his discussion of autocorrelation of datapoints before regime shifts as being innovative, but in this case while I do think they provide a good way of discussing how dire (perhaps hopeless) environmental and social conditions are in the world right now(even in the Netherlands?!), this piece doesn't seem to groundbreaking. Others? ~Sam

Scheffer way to see interaction between social and ecological system is based on the contradiction. In other words increase in human welfare means increase in ecosystem stress. Along their line of argumentation about implications of social choices of using ecosystem resources they relay in a old style argumentation of how to deal with power, putting the state as able to design and implement policies to get the best deal for all stakeholders. The second point that I would like to highlight is the importance of good understanding about the ecosystem dynamics to guide political decision out of the edge. The argumentation is that if we consider the size of the country economy, decision over a localized ecosystem could be done searching for the public optimum. But as we see in the Amazon and elsewhere more, political and economic drivers are drove by direct users (Affectors in paper language). I really like the idea of graphic representation, although as Sam present, bring not new to the debate Ricmel

Is it possible to facilitate transformation? During the conference, I targeted presentations on “learning” and “social networks.” Several of these discussed participatory modeling, multi-stakeholder management of resources, and the facilitation of visioning processes. However, they often departed from a “tool-box” approach that incorporated simulation games or similar techniques. I believe that this approach fails to emphasize the social conditions of engagement, such as the complexities inherent in navigating diverse stakeholder needs, the various motivations that drive participation, or the implications of unbalanced knowledge, skills, or power. The situations/contexts in which these games are used deserve far greater attention. In a few cases presenters discussed how to improve the process of learning among stakeholders and design conditions under which social learning is more likely to occur. However, in general, research projects seemed to be guided by the assumption that learning and participation simply happen when you bring people together around the table. Little sensitivity was given to how people participate, how differences and tensions are managed, or how alliances form. More effort is needed to describe and evaluate how participation, reflection and learning among stakeholders impact outcomes and relationships. I was relieved when I heard Bill Clark from Harvard admit that “Imagining a future is very different from capturing it. Learning is a political act and so the question becomes - Who gets to the table?” Network presentations emphasized that one might manipulate networks by designing interventions that build ties, by seeking leverage points in the system, or by creating networks that lead to emergent entrepreneurial activity. In a world of contested power and conflicting interests I was left with the following questions: Who might be deemed appropriate to conduct all this manipulation? How does the selection of these decision makers affect the potential to transform social systems vs. maintain the status quo? WL


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User talk:Cjr100B

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Links
Hi, did you know that dog stays for dog? If you are using a features from betaversion you have there a bottom for this.--Juan de Vojníkov 10:14, 17 July 2011 (UTC)

Federal Writers' Project - Life Histories
Hi! Excellent work on the Federal Writers' Project - Life Histories. I like the quality of the content your students are required to submit.

I wanted to let you know that we're trying to better organize the content on Wikiversity so that pages are saved with the learning project they are associated with. For this project, all of the life histories have been moved under the project with links such as Federal Writers' Project - Life Histories/Norman T. Banks. This improves navigation by allowing easy movement between subpages and back to the project. The project itself has been updated so that all links work correctly with the new page locations.

If you have students add new life histories to the project in the future, please update your instructions to have the students save files under the learning project with a title / link similar to the one above. Let me know if you have any questions. Thanks!

Dave Braunschweig (discuss • contribs) 03:24, 2 January 2014 (UTC)

title: User talk:Cjr100B


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text:
Social welfare

Defining Social Welfare
While the word welfare is often used in relation to an individual's state of dependence on government services. Social welfare, simply put, relates to the wellness of communities and the larger society. Members of communities gauge their wellness related to whether or not they have jobs, housing, safety, relationships, etc. Social welfare (community wellness) is a subjective term, as there are varying opinions about what constitutes a "well" society. Therefore, the term becomes a blank slate where individuals' ideology and value positions give definition and ultimately leave the term open for individual interpretation.

title: Social welfare


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User:Albstp

Am louis frm Ghana. I like french and hope to master it with your help.

title: User:Albstp


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text:
Engineering Experience 4: Design a Small Solar Vehicle/Nl/Team PM7/Design rapport/genetisch algoritme

Uitleg
Aangezien er 3 overbrengingen (per versnelling) zijn, zijn er 6 tandwielen nodig per versnelling. Er zijn 2 overbrengingen gemeenschappelijk, terwijl een 3e de 2 versnellingen differentieert. In totaal zijn er dus 8 tandwielen nodig. In de catalogus zijn er 30 soorten tandwielen die bruikbaar zijn voor de zonnewagen. Het aantal mogelijke combinaties is dus 30^8 of 656 100 000 000. De naïeve manier van alle combinaties te overlopen zou vrij veel tijd in beslag nemen.

Dankzij een techniek uit de artificiële intelligentie kan de oplossing sneller bekomen worden, of op zijn minst benadert worden. Deze techniek die een beetje gelijkt op wat er bij voortplanting met genen gebeurt wordt een genetisch algoritme genoemd.

Bij genetische algoritmen, wordt een oplossing voorgesteld door een set van waarden (genoom). Deze set heeft ook een "fitness" die weergeeft hoe ver van een ideale oplossing deze combinatie ligt. In dit geval, bevat stellen de waarden in de set de tandwielen voor. Er zijn dus 8 waarden. De fitness geeft weer hoe ver de oplossing van de 2 gevraagde overbrengingsverhoudingen ligt.

Verder is er een "pool" van oplossingen. In dit geval worden de 20 beste oplossingen bij gehouden. Uit deze pool worden 2 oplossingen genomen en op een willekeurige plaats gekruist. Stel dat een oplossing wordt voorgesteld door AAAAAAAA en een andere door BBBBBBBB. Een kruising op de 3e positie geeft AABBBBBB en BBAAAAAA terug.

Vervolgens kunnen er ook nog mutaties gebeuren. In dit geval is een mutatie de vergroting of verkleining van een van de tandwielen in de oplossing.

Na de kruising en mutatie wordt de fitness van de nieuwe oplossing berekent en toegevoegd aan de pool als hij bij de beste oplossingen hoort.

Broncode
[Maple broncode]


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User:Eml5526.s11.team7/HW5

Problem Description

 * 1) Solve the general 1-D model for an elastic bar with data set 1b (G1DM1.0/D1b) using the weight function with polynomial, Fourier, and exponential basis functions until convergence of $$\displaystyle u^h (0.5) \rightarrow O(10^{-6}) $$.
 * 2) Plot u, uh, and convergence (error versus n).

Given
For G1DM1.0/D1 and D1b, the governing partial differential equation (PDE) and given conditions are,

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

$$\displaystyle \frac{\partial}{\partial x} \left[ a_2(x) \frac{\partial u}{\partial x} \right ] + f(x,t) = \bar{m} \frac{\partial ^s u}{\partial t ^s} (x,t)$$ (5.1.1)
 * style="width:10%; |
 * style="width:10%; |
 *
 * }

 * where
 * $$\displaystyle \Omega = [{0,1}]$$

 * $$\displaystyle a_2 (x) = 2 + 3x$$

 * $$\displaystyle f(x) = 5x$$

 * $$\displaystyle \bar{m} \frac{\partial ^s u}{\partial t ^s} (x,t) = 0 $$ (static case)

The natural and essential boundary conditions are respectively defined below for data set D1b.

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

$$ n(1) a_2(1) \frac{\partial u}{\partial x}(1)=12 \qquad \Rightarrow \qquad \frac{\partial u}{\partial x}(1)= \frac{12}{5}$$ (5.1.2)
 * style="width:10%; |
 * style="width:10%; |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle u(0)=4$$ (5.1.3)
 * style="width:10%; |
 * style="width:10%; |
 *
 * }

Weak Solution
The weak form must first be constructed such that the weight function, $$\displaystyle w(x)$$, vanishes on the essential boundary (x=0). First, the governing equation (5.1.1) and natural boundary condition (5.1.2) are multiplied by the weight function and integrated over the domain. The restriction on the weight function is that $$\displaystyle w(0)=0$$.

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

$$ \int_0^1 w(x) \left[ \frac{d}{dx} \left(a_2(x) \frac{du}{dx} \right) + f(x) \right] dx = 0 \qquad and \qquad w(x)\left( \frac{du}{dx} - \frac{12}{5} \right )_{x=1} = 0 $$     (5.1.4)
 * style="width:10%; |
 * style="width:10%; |
 *
 * }

The first equation can be expanded as

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

$$ \int_0^1 w(x) \left[ \frac{d}{dx} \left(a_2(x) \frac{du}{dx} \right) \right] dx + \int_0^1 w(x) f(x) dx= 0 \qquad \forall w. $$ (5.1.5)
 * style="width:10%; |
 * style="width:10%; |
 *
 * }

Integrating by parts yields

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

$$ \int_0^1 w(x) \frac{d}{dx} \left(a_2(x) \frac{du}{dx} \right) dx = \left( w(x)a_2(x) \frac{du}{dx} \right )_{x=1} - \left( w(x)a_2(x) \frac{du}{dx} \right )_{x=0} - \int_0^1 \frac{dw}{dx}a_2(x) \frac{du}{dx} dx. $$     (5.1.6)
 * style="width:10%; |
 * style="width:10%; |
 *
 * }

Recall that weight function must vanish at the essential boundary. Then by combining equations 5.1.5 and 5.1.6, the following expression is determined.

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

$$ \int_0^1 \frac{dw}{dx}a_2(x) \frac{du}{dx} dx = \left( w(x) a_2(x) \frac{du}{dx} \right)_{x=1} + \int_0^1 w(x)f(x)dx \qquad \forall w \quad with \quad w(0)=0. $$     (5.1.7)
 * style="width:10%; |
 * style="width:10%; |
 *
 * }

Then substitute the natural boundary condition (equation 5.1.2) and the given conditions for $$\displaystyle a_2(x)$$ and $$\displaystyle f(x)$$ into the previous equation. The result is the weak form for G1DM1.0/D1b.

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

$$ \int_0^1 \frac{dw}{dx}(2+3x) \frac{du}{dx} dx = 12w(1) + \int_0^1 w(x)5x dx \qquad \forall w \quad with \quad w(0)=0. $$     (5.1.8)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 *
 * }

Exact Solution
The governing partial differential equation (PDE):
 * {| style="width:100%" border="0"

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

\frac{\partial }\left[ {\left( {2 + 3x} \right)\frac} \right] + 5x = 0{\text{ }}\forall {\text{x}} \in \left] {0,1} \right[

$$     (5.1.9) The natural and essential B.C.:
 *
 * }
 * {| style="width:100%" border="0"

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

\frac\left( 1 \right) = \frac{5}

$$     (5.1.10)
 *
 * }
 * {| style="width:100%" border="0"

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

u\left( 0 \right) = 4

$$     (5.1.11) Then
 *
 * }
 * {| style="width:100%" border="0"

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

\int {\frac{\partial }\left[ {\left( {2 + 3x} \right)\frac} \right]} dx = \int { - 5xdx}

$$     (5.1.12)
 *
 * }
 * {| style="width:100%" border="0"

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

\left( {2 + 3x} \right)\frac + {C_1} = - \frac{5}{2}{x^2}{\text{ where }}{C_1}{\text{ is a constant}}

$$     (5.1.13) By the natural B.C.
 *
 * }
 * {| style="width:100%" border="0"

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

{C_1} = - \frac{2}

$$     (5.1.14) Then
 *
 * }
 * {| style="width:100%" border="0"

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

\frac = \frac

$$     (5.1.15)
 *
 * }
 * {| style="width:100%" border="0"

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

u = - \frac{5}{x^2} + \frac{5}{9}x + \frac\log \left( {3x + 2} \right) + {C_2}{\text{ where }}{C_2}{\text{ is a constant}}

$$     (5.1.16) By the essential B.C.
 *
 * }
 * {| style="width:100%" border="0"

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

{C_2} = 4 - \frac\log \left( 2 \right)

$$     (5.1.17) Then
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

u = - \frac{5}{x^2} + \frac{5}{9}x + \frac\log \left( {3x + 2} \right) + 4 - \frac\log \left( 2 \right)

$$     (5.1.18)
 *

 * }

For Polynomial Basis
The corresponding family $${{\bar{F}}_{P}}=\left\{ {{{\bar{b}}}_{j}} \right\}$$ of polynomial basis function satisfying the Constraint Breaking Solution (CBS) for $$\Gamma_g = \left \{ 0 \right \}$$ is:

$${{F}_{P}}=\left\{ {{\left( x \right)}^{j}},\text{ }j=0,1,2,\ldots \right\}$$

Now that the family of basis functions are determined, we recall that the discrete weak form is defined by

$$ \mathbf{\tilde{d}} \cdot \mathbf{b}( \beta ) = g \qquad \text{and} \qquad \mathbf{\tilde{c}} \cdot \left[ \mathbf{\tilde{M}} \mathbf{\tilde{d}}^{(s)} + \mathbf{\tilde{K}} \mathbf{\tilde{d}} - \mathbf{\tilde{F}} \right] = 0 $$

where the expression $$\displaystyle \mathbf{\tilde{M}} \mathbf{\tilde{d}}^{(s)} = 0 $$ for the static case. Then the discrete weak form is simplified to:

\mathbf{K} \mathbf{d} = \mathbf{F} \qquad \Rightarrow \qquad \mathbf{K} \mathbf{d} = \mathbf{F}_F - \mathbf{K}_{FE} g. $$

The first constant, $$\displaystyle d_0$$, is determined from the formulation of the CBS. It is known that:

 * $$\displaystyle u^h(\beta) = d_0 b_0(\beta)=g$$

 * $$\displaystyle d_0= \frac{g}{b_0(\beta)}=4/1=4$$

Now the matrices used to determine the rest of the coefficients are determined as follows. Note that the number of terms needed in the approximate solution, $$\displaystyle n$$, is determined by the convergence interval given. The first few terms of each matrix computation is shown. A Matlab script was written to iterate these computation for increasing values of $$\displaystyle n$$ until the desired accuracy was reached.

\displaystyle

{F_F} = \left\{ {{F_i},i = 1,2,...,n} \right\} = \left\{ {\tilde f\left( \right),i = 1,2,...,n} \right\}

$$

\displaystyle

\tilde f({b_i}) = {b_i}(\alpha )h + \int\limits_\alpha ^\beta {{b_i}fdx}

$$

\displaystyle

h = {a_2}(\alpha )\frac = (2+3\alpha)\frac = 5*12/5=12

$$

\displaystyle

\tilde f({b_1}) = {b_1}(1)h + \int\limits_0^1 {{b_1}(5x)dx}=1*12+\int\limits_0^1  {{(x)}(5x)dx}=41/3

$$

\displaystyle

\tilde f({b_2}) = {b_2}(1)h + \int\limits_0^1 {{b_2}(5x)dx}=1*12 + \int\limits_0^1  {{(x)^2}(5x)dx}=53/4

$$

$$ \displaystyle

\therefore {F_F} = \left[ {\begin{array}{ccccccccccccccc} {41/3} \\ {53/4} \\  {.} \\  {.} \\  {.} \end{array}} \right]

$$

\displaystyle

{K_{FE}} = \left\{ {{K_{0j}};j = 1,...,n} \right\} = \left\{ {\tilde k\left( {{b_0},{b_j}} \right),j = 1,...,n} \right\}

$$

\displaystyle

\tilde k({b_i},{b_j}) = \int\limits_0^1 {b{'_i}{a_2}b{'_j}dx}

$$

\displaystyle

\tilde k({b_0},{b_1}) = \int\limits_0^1 {b{'_0}{a_2}b{'_1}dx} = \int\limits_0^1 {b{'_0}(2+3x)b{'_1}dx}  = 0 = \tilde k({b_0},{b_2}) = \tilde k({b_0},{b_3}) = 0

$$

\displaystyle

\therefore {K_{EF}} = \left[ {\begin{array}{ccccccccccccccc} 0 \\ 0 \\  . \\ .\\ . \end{array}} \right]

$$

And：

\displaystyle

\because {K_{FE}} = {K_{EF}}^T = \left[ {\begin{array}{ccccccccccccccc} 0 \\ 0 \\  . \\ .\\ . \end{array}} \right]

$$

\displaystyle

{K_{FF}} = \left[ {{K_{ij}};i,j = 1,2,...} \right] = \tilde k({b_i},{b_j})

$$

\displaystyle

\tilde k({b_1},{b_1}) = \int\limits_0^1 {b{'_1}{2+3x}b{'_1}dx} = \int\limits_0^1 {1*(2+3x)*1dx}=7/2

$$

\displaystyle

\tilde k({b_1},{b_2}) = \int\limits_0^1 {b{'_1}{2+3x}b{'_2}dx} = \int\limits_0^1 {1*(2+3x)*2(x)dx}=4

$$

\displaystyle

\tilde k({b_2},{b_1}) = \int\limits_0^1 {b{'_2}{2+3x}b{'_1}dx} = \int\limits_0^1 {2(x)*(2+3x)*1dx}=4

$$

\displaystyle

\tilde k({b_2},{b_2}) = \int\limits_0^1 {b{'_2}{2+3x}b{'_2}dx} = \int\limits_0^1 {2(x)*(2+3x)*2(x)dx}=17/3

$$

\displaystyle \therefore {K_{FF}} = \begin{bmatrix} 7/2&4 &.&. \\ 4&17/3  & .&.\\ . &.  & .&.\\ . &.  & .&. \end{bmatrix} $$

The matrices are then solved for the coefficients:

\displaystyle

Kd = F $$

\displaystyle

F = {F_F} - K_{FE}g = {F_F} $$

\displaystyle

K_{FF}d_F=F_F $$

\displaystyle

d_F=K_{FF}^{-1} F_F $$

The approximate solution, $$\displaystyle u^h(x)$$ is defined as,

 * $$\displaystyle u^h(x)= d_0b_0+\sum_{j=1}^{n}d_jb_j(x) \qquad \text{where} \quad d_0=4 \ \text{and} \ b_0=1$$.

As stated above a Matlab script was written to iterate $$\displaystyle n$$ until the desired accuracy was reached. At $$\displaystyle n=6$$ the error was $$\displaystyle  O(3.5840*10^{-6})$$. The following plots were generated:

For Fourier Basis
The family of Fourier basis functions is of the form
 * $$\displaystyle F_F = \left\{ 1, cos(x-1), sin(x), cos(2x)-1, sin(2x),..., cos\left( \frac{nx}{2}\right)-1, sin\left(\frac{nx}{2}\right) \right\}$$.

Now that the family of basis functions are determined, we recall that the discrete weak form is defined by

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

$$ \mathbf{\tilde{d}} \cdot \mathbf{b}( \beta ) = g \qquad \text{and} \qquad \mathbf{\tilde{c}} \cdot \left[ \mathbf{\tilde{M}} \mathbf{\tilde{d}}^{(s)} + \mathbf{\tilde{K}} \mathbf{\tilde{d}} - \mathbf{\tilde{F}} \right] = 0 $$
 * style="width:10%; |
 * style="width:10%; |
 * }

 * where the expression $$\displaystyle \mathbf{\tilde{M}} \mathbf{\tilde{d}}^{(s)} = 0 $$ for the static case. Then equation 5.1.11 for the unconstrained case can be rewritten as

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

$$ \mathbf{K} \mathbf{d} = \mathbf{F} \qquad \Rightarrow \qquad \mathbf{K} \mathbf{d} = \mathbf{F}_F - \mathbf{K}_{FE} g. $$
 * style="width:10%; |
 * style="width:10%; |
 * }

This agrees with Vu-Quoc's notes. Now the matrices must be determined. From the derivation of the polynomial basis case above, it holds that $$\displaystyle d_0 = 4 $$, and $$\displaystyle \mathbf{d}_F = \mathbf{K}_{FF}^{-1} \mathbf{F}_F$$.

 * For n=6,

 * $$\displaystyle

\mathbf{F}_F = \tilde f(b_i) = {b_i}(1)h + \int_0^1 {{b_i}(5x)dx} $$

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

$$ \mathbf{F}_F = \begin{bmatrix} \tilde f(b_1) \\ \tilde f(b_2) \\ \tilde f(b_3) \\ \tilde f(b_4) \\ \tilde f(b_5) \\ \tilde f(b_6) \end{bmatrix} \qquad \Rightarrow \qquad \mathbf{F}_F = \begin{bmatrix} -6.108 \\ 11.603 \\ -18.991 \\ -13.089 \\ -27.25 \\ 3.422 \end{bmatrix} $$
 * style="width:10%; |
 * style="width:10%; |
 * }

The stiffness matrix is determined,

 * $$\displaystyle

\mathbf{K}_{FF} = \tilde k(b_i,b_j) = \int_0^1 {b_i}'(2+3x){b_j}'dx $$

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

$$ \mathbf{K}_{FF} = \begin{bmatrix} 1.144 & -1.361 & 3.256 & -0.541 & 3.701 & 2.321 \\ -1.361 & 2.356 & -4.187 & 2.4 & -5.846 & -0.067 \\ 3.256 & -4.187 & 9.512 & -2.35 & 11.619 & 5.465 \\ -0.541 & 2.4 & -2.35 & 4.488 & -5.473 & 5.35 \\ 3.701 & -5.846 & 11.619 & -5.473 & 16.813 & 2.205 \\ 2.321 & -0.067 & 5.465 & 5.35 & 2.205 & 14.687 \end{bmatrix} $$
 * style="width:10%; |
 * style="width:10%; |
 * }

Now, the matrix $$\displaystyle \mathbf{d}_F$$ can be found.

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

$$ \mathbf{d}_{F} = \mathbf{K}_{FF}^{-1} \mathbf{F}_{F} \qquad \Rightarrow \qquad \mathbf{d}_{F} = \begin{bmatrix} 17.629 \\ 23.592 \\ 0.383 \\ -9.94 \\ -0.952 \\ 1.177 \end{bmatrix} $$
 * style="width:10%; |
 * style="width:10%; |
 * }

The approximate solution, $$\displaystyle u^h(x)$$ is defined as,

 * $$\displaystyle u^h(x)= d_0b_0+\sum_{j=1}^{n}d_jb_j(x) \qquad \text{where} \quad d_0=4 \ \text{and} \ b_0=1$$.

 * OR (simplified)

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

$$\displaystyle u^h(x)= 17.629*cos(x) \ + \ 23.592*sin(x) \ + \ 0.383*cos(2x) \ - \ 9.94*sin(2x) \ - \ 0.952*cos(3x) \ + \ 1.177*sin(3x) \ - \ 13.06 $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * }

For Exponential Basis
The exponential basis begins to converge as n increases to around n=3, but then quickly begins to diverge. At n=10, the solution begins to diverge at the right end, and when n=20, it is clear that the solution is not stable.

Problem Description
Rework the previous problem for data set 1, i.e. G1DM1.0/D1.
 * Note: The difference between data set 1 and data set 1b is where the boundary conditions are applied. Specifically, the essential boundary condition for D1 is applied at x=1 while the natural boundary condition is applied at x=0. The opposite is true for data set 1b.

The essential and natural boundary conditions are respectively defined below for data set D1.
 * $$ \Gamma_g = \left \{ 1 \right \} \quad \text{and} \quad \Gamma_h = \left \{ 0 \right \}$$

 * $$ n(0) a_2(0) \frac{\partial u}{\partial x}(0)=-12 \qquad \Rightarrow \qquad \frac{\partial u}{\partial x}(0)= 6$$
 * $$\displaystyle u(1)=4$$

Weak Solution
The weak form must first be constructed such that the weight function, $$\displaystyle w(x)$$, vanishes on the essential boundary (x=0). First, the governing equation (5.1.1) and natural boundary condition (5.1.2) are multiplied by the weight function and integrated over the domain. The restriction on the weight function is that $$\displaystyle w(1)=0$$.

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

$$ \int_0^1 w(x) \left[ \frac{d}{dx} \left(a_2(x) \frac{du}{dx} \right) + f(x) \right] dx = 0 \qquad and \qquad w(x)\left( \frac{du}{dx} - \frac{12}{5} \right )_{x=0} = 0 $$     (5.2.1)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

The first equation can be expanded as

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

$$ \int_0^1 w(x) \left[ \frac{d}{dx} \left(a_2(x) \frac{du}{dx} \right) \right] dx + \int_0^1 w(x) f(x) dx= 0 \qquad \forall w. $$ (5.2.2)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

Integrating by parts yields

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

$$ \int_0^1 w(x) \frac{d}{dx} \left(a_2(x) \frac{du}{dx} \right) dx = \left( w(x)a_2(x) \frac{du}{dx} \right )_{x=1} - \left( w(x)a_2(x) \frac{du}{dx} \right )_{x=0} - \int_0^1 \frac{dw}{dx}a_2(x) \frac{du}{dx} dx. $$     (5.2.3)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

Recall that weight function must vanish at the essential boundary. Then by combining equations 5.1.5 and 5.1.6, the following expression is determined.

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

$$ \int_0^1 \frac{dw}{dx}a_2(x) \frac{du}{dx} dx =- \left( w(x) a_2(x) \frac{du}{dx} \right)_{x=0} + \int_0^1 w(x)f(x)dx \qquad \forall w \quad with \quad w(0)=0. $$     (5.2.4)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

Then substitute the natural boundary condition (equation 5.1.2) and the given conditions for $$\displaystyle a_2(x)$$ and $$\displaystyle f(x)$$ into the previous equation. The result is the weak form for G1DM1.0/D1.

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

$$ \int_0^1 \frac{dw}{dx}(2+3x) \frac{du}{dx} dx = -12w(0) + \int_0^1 w(x)5x dx \qquad \forall w \quad with \quad w(0)=0. $$     (5.2.5)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Exact Solution
The governing partial differential equation (PDE):
 * {| style="width:100%" border="0"

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

\frac{\partial }\left[ {\left( {2 + 3x} \right)\frac} \right] + 5x = 0{\text{ }}\forall {\text{x}} \in \left] {0,1} \right[

$$     (5.2.6) The natural and essential B.C.:
 *

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

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

\frac\left( 0 \right) = 6

$$     (5.2.7)
 *

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

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

u\left( 1 \right) = 4

$$     (5.2.8) Then
 *

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

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

\int {\frac{\partial }\left[ {\left( {2 + 3x} \right)\frac} \right]} dx = \int { - 5xdx}

$$     (5.2.9)
 *

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

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

\left( {2 + 3x} \right)\frac + {C_1} = - \frac{5}{2}{x^2}{\text{ where }}{C_1}{\text{ is a constant}}

$$     (5.2.10) By the natural B.C.
 *

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

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

{C_1} = - 12

$$     (5.2.11) Then
 *

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

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

\frac = \frac

$$     (5.2.12)
 *

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

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

u = - \frac{5}{x^2} + \frac{5}{9}x - \frac\log \left( {3x + 2} \right) + \frac{5}{9} + {C_2}{\text{ where }}{C_2}{\text{ is a constant}}

$$     (5.2.13) By the essential B.C.
 *

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

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

{C_2} = 4 + \frac\log \left( 5 \right)-\frac{10}{9}+\frac{5}{12]}

$$     (5.2.14) Then
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

u = - \frac{5}{x^2} + \frac{5}{9}x - \frac\log \left( {3x + 2} \right) + \frac{5}{12} + \frac\log \left( 5 \right)+4-\frac{5}{9}

$$     (5.2.15)
 *

 * }

For Polynomial Basis
The corresponding family $${{\bar{F}}_{P}}=\left\{ {{{\bar{b}}}_{j}} \right\}$$ of polynomial basis function satisfying the Constraint Breaking Solution (CBS) for $$\Gamma_g = \left \{ 1 \right \}$$ is:

$${{F}_{P}}=\left\{ {{\left( x-1 \right)}^{j}},\text{ }j=0,1,2,\ldots \right\}$$

Now that the family of basis functions are determined, we recall that the discrete weak form is defined by

$$ \mathbf{\tilde{d}} \cdot \mathbf{b}( \beta ) = g \qquad \text{and} \qquad \mathbf{\tilde{c}} \cdot \left[ \mathbf{\tilde{M}} \mathbf{\tilde{d}}^{(s)} + \mathbf{\tilde{K}} \mathbf{\tilde{d}} - \mathbf{\tilde{F}} \right] = 0 $$

where the expression $$\displaystyle \mathbf{\tilde{M}} \mathbf{\tilde{d}}^{(s)} = 0 $$ for the static case. Then the discrete weak form is simplified to:

\mathbf{K} \mathbf{d} = \mathbf{F} \qquad \Rightarrow \qquad \mathbf{K} \mathbf{d} = \mathbf{F}_F - \mathbf{K}_{FE} g. $$

The first constant, $$\displaystyle d_0$$, is determined from the formulation of the CBS. It is known that:

 * $$\displaystyle u^h(\beta) = d_0 b_0(\beta)=g$$

 * $$\displaystyle d_0= \frac{g}{b_0(\beta)}=4/1=4$$

Now the matrices used to determine the rest of the coefficients are determined as follows. Note that the number of terms needed in the approximate solution, $$\displaystyle n$$, is determined by the convergence interval given. The first few terms of each matrix computation is shown. A Matlab script was written to iterate these computation for increasing values of $$\displaystyle n$$ until the desired accuracy was reached.

\displaystyle

{F_F} = \left\{ {{F_i},i = 1,2,...,n} \right\} = \left\{ {\tilde f\left( \right),i = 1,2,...,n} \right\}

$$

\displaystyle

\tilde f({b_i}) = {b_i}(\alpha )h + \int\limits_\alpha ^\beta {{b_i}fdx}

$$

\displaystyle

h = {a_2}(\alpha )\frac = (2+3\alpha)\frac = 2*6=12

$$

\displaystyle

\tilde f({b_1}) = {b_1}(0)h + \int\limits_0^1 {{b_1}(5x)dx}=1*12+\int\limits_0^1  {{(x-1)}(5x)dx}=-77/6

$$

\displaystyle

\tilde f({b_2}) = {b_2}(0)h + \int\limits_0^1 {{b_2}(5x)dx}=-1*12 + \int\limits_0^1  {{(x-1)^2}(5x)dx}=149/12

$$

$$ \displaystyle

\therefore {F_F} = \left[ {\begin{array}{ccccccccccccccc} {-77/6} \\ {149/12} \\  {.} \\  {.} \\  {.} \end{array}} \right]

$$

\displaystyle

{K_{FE}} = \left\{ {{K_{0j}};j = 1,...,n} \right\} = \left\{ {\tilde k\left( {{b_0},{b_j}} \right),j = 1,...,n} \right\}

$$

\displaystyle

\tilde k({b_i},{b_j}) = \int\limits_0^1 {b{'_i}{a_2}b{'_j}dx}

$$

\displaystyle

\tilde k({b_0},{b_1}) = \int\limits_0^1 {b{'_0}{a_2}b{'_1}dx} = \int\limits_0^1 {b{'_0}(2+3x)b{'_1}dx}  = 0 = \tilde k({b_0},{b_2}) = \tilde k({b_0},{b_3}) = 0

$$

\displaystyle

\therefore {K_{EF}} = \left[ {\begin{array}{ccccccccccccccc} 0 \\ 0 \\  . \\ .\\ . \end{array}} \right]

$$

And：

\displaystyle

\because {K_{FE}} = {K_{EF}}^T = \left[ {\begin{array}{ccccccccccccccc} 0 \\ 0 \\  . \\ .\\ . \end{array}} \right]

$$

\displaystyle

{K_{FF}} = \left[ {{K_{ij}};i,j = 1,2,...} \right] = \tilde k({b_i},{b_j})

$$

\displaystyle

\tilde k({b_1},{b_1}) = \int\limits_0^1 {b{'_1}{2+3x}b{'_1}dx} = \int\limits_0^1 {1*(2+3x)*1dx}=7/2

$$

\displaystyle

\tilde k({b_1},{b_2}) = \int\limits_0^1 {b{'_1}{2+3x}b{'_2}dx} = \int\limits_0^1 {1*(2+3x)*2(x-1)dx}=-3

$$

\displaystyle

\tilde k({b_2},{b_1}) = \int\limits_0^1 {b{'_2}{2+3x}b{'_1}dx} = \int\limits_0^1 {2(x-1)*(2+3x)*1dx}=-3

$$

\displaystyle

\tilde k({b_2},{b_2}) = \int\limits_0^1 {b{'_2}{2+3x}b{'_2}dx} = \int\limits_0^1 {2(x-1)*(2+3x)*2(x-1)dx}=11/3

$$

\displaystyle \therefore {K_{FF}} = \begin{bmatrix} 7/2&-3 &.&. \\ -3&11/3  & .&.\\ . &.  & .&.\\ . &.  & .&. \end{bmatrix} $$

The matrices are then solved for the coefficients:

\displaystyle

Kd = F $$

\displaystyle

F = {F_F} - K_{FE}g = {F_F} $$

\displaystyle

K_{FF}d_F=F_F $$

\displaystyle

d_F=K_{FF}^{-1} F_F $$

The approximate solution, $$\displaystyle u^h(x)$$ is defined as,

 * $$\displaystyle u^h(x)= d_0b_0+\sum_{j=1}^{n}d_jb_j(x) \qquad \text{where} \quad d_0=4 \ \text{and} \ b_0=1$$.

As stated above a Matlab script was written to iterate $$\displaystyle n$$ until the desired accuracy was reached. At $$\displaystyle n=6$$ the error was $$\displaystyle  O(3.4091*10^{-6})$$. The following plots were generated:

For Fourier Basis
The family of Fourier basis functions is of the form
 * $$\displaystyle F_F = \left\{ 1, cos(x-1)-1, sin(x-1), cos[2(x-1)]-1, sin[2(x-1)],..., cos[\frac{n}{2}(x-1)]-1, sin[\frac{n}{2}(x-1) \right\}$$.

Now that the family of basis functions are determined, we recall that the discrete weak form is defined by

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

$$ \mathbf{\tilde{d}} \cdot \mathbf{b}( \beta ) = g \qquad \text{and} \qquad \mathbf{\tilde{c}} \cdot \left[ \mathbf{\tilde{M}} \mathbf{\tilde{d}}^{(s)} + \mathbf{\tilde{K}} \mathbf{\tilde{d}} - \mathbf{\tilde{F}} \right] = 0 $$
 * style="width:10%; |
 * style="width:10%; |
 * }

 * where the expression $$\displaystyle \mathbf{\tilde{M}} \mathbf{\tilde{d}}^{(s)} = 0 $$ for the static case. Then equation 5.1.11 for the unconstrained case can be rewritten as

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

$$ \mathbf{K} \mathbf{d} = \mathbf{F} \qquad \Rightarrow \qquad \mathbf{K} \mathbf{d} = \mathbf{F}_F - \mathbf{K}_{FE} g. $$
 * style="width:10%; |
 * style="width:10%; |
 * }

This agrees with Vu-Quoc's notes. Now the matrices must be determined. From the derivation of the polynomial basis case above, it holds that $$\displaystyle d_0 = 4 $$, and $$\displaystyle \mathbf{d}_F = \mathbf{K}_{FF}^{-1} \mathbf{F}_F$$.

 * For n=6,

 * $$\displaystyle

\mathbf{F}_F = \tilde f(b_i) = {b_i}(0)h + \int_0^1 {{b_i}(5x)dx} $$

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

$$ \mathbf{F}_F = \begin{bmatrix} \tilde f(b_1) \\ \tilde f(b_2) \\ \tilde f(b_3) \\ \tilde f(b_4) \\ \tilde f(b_5) \\ \tilde f(b_6) \end{bmatrix} \qquad \Rightarrow \qquad \mathbf{F}_F = \begin{bmatrix} -5.718 \\ -10.89 \\ -17.724 \\ -12.275 \\ -25.274 \\ -3.282 \end{bmatrix} $$
 * style="width:10%; |
 * style="width:10%; |
 * }

The stiffness matrix is determined,

 * $$\displaystyle

\mathbf{K}_{FF} = \tilde k(b_i,b_j) = \int_0^1 {b_i}'(2+3x){b_j}'dx $$

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

$$ \mathbf{K}_{FF} = \begin{bmatrix} 0.764 & 1.117 & 2.305 & 0.885	& 3.06	& -0.773 \\ 1.117 & 2.736 & 3.675 & 3.819 & 5.929 & 2.855 \\ 2.305 & 3.675 & 7.137 & 3.437 & 10.079 & -1.18 \\ 0.885 & 3.819 & 3.437 & 6.863 & 7.19 & 8.293 \\ 3.06 & 5.929 & 10.079 & 7.19 & 16.154 & 2.415 \\ -0.773 & 2.855 & -1.18 & 8.293 & 2.415 & 15.346 \end{bmatrix} $$
 * style="width:10%; |
 * style="width:10%; |
 * }

Now, the matrix $$\displaystyle \mathbf{d}_F$$ can be found.

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

$$ \mathbf{d}_{F} = \mathbf{K}_{FF}^{-1} \mathbf{F}_{F} \qquad \Rightarrow \qquad \mathbf{d}_{F} = \begin{bmatrix} -26.379 \\ 1.214 \\ 8.796 \\ -3.542 \\ -1.074 \\ 0.991 \end{bmatrix} $$
 * style="width:10%; |
 * style="width:10%; |
 * }

The approximate solution, $$\displaystyle u^h(x)$$ is defined as,

 * $$\displaystyle u^h(x)= d_0b_0+\sum_{j=1}^{n}d_jb_j(x) \qquad \text{where} \quad d_0=4 \ \text{and} \ b_0=1$$.

 * OR (simplified)

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

$$\displaystyle u^h(x)= - 26.379*cos(x-1) \ + \ 1.214*sin(x-1) \ + \ 8.796*cos(2x-2) \ - \ 3.542*sin(2x-2)] \ - \ 1.074*cos(3x-3) \ + \ 0.991*sin(3x-3) \ + \ 22.657 $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * }

For Exponential Basis
The exponential basis function is given by:
 * {| style="width:100%" border="0"

$$ \displaystyle {u^{h}}(x)={d_{j}}\left (e^{j(x-1)}-1 \right ) $$       (5.2.exp.1)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

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

$$ \displaystyle \frac{{u^{h}}(x)}{dx} ={d_{j}}\left (e^{j(x-1)} \right ) $$       (5.2.exp.2) The general form of the stiffness matrix becomes:
 * style="width:10%; |
 * style="width:10%; |
 *

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

$$ \displaystyle K=\int_{0}^{1}i*e^{i(x-1)}*(2+3x)*j*e^{j(x-1)}dx $$       (5.2.exp.3) The general form of the force matrix becomes:
 * style="width:10%; |
 * style="width:10%; |
 *

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

$$ \displaystyle F=12(e^{i(0-1)}-1)+\int_{0}^{1} 5x*i(e^{i(x-1)}-1)dx $$       (5.2.exp.4) Using MatLab, the K and F matrix with N=10 is:
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

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

$$ \displaystyle K=\begin{bmatrix} 1.716166179 &	2.633475288	& 3.170329089	& 3.512453499	&3.74690156	&3.917139098	&4.046398014	&4.147965319	&4.229930538	&4.297494435\\ 2.633475288	&4.227105451	&5.268680249	&5.995042496	&6.528565163	&6.93668231	&7.258939308	&7.519876512	&7.735489984	&7.916648746\\ 3.170329089	&5.268680249	&6.744422808	&7.834278196	&8.670852887	&9.332921967	&9.869837922	&10.31398664	&10.68747581	&11.00590793\\ 3.512453499	&5.995042496	&7.834278196	&9.248909746	&10.3699133	&11.27981477	&12.03298442	&12.66663799	&13.20708952	&13.67346515\\ 3.74690156	&6.528565163	&8.670852887	&10.3699133	&11.74980705	&12.89248331	&13.85413531	&14.67454391	&15.38264829	&15.99999816\\ 3.917139098	&6.93668231	&9.332921967	&11.27981477	&12.89248331	&14.24996774	&15.4082711	&16.40815817	&17.27999802	&18.04687424\\ 4.046398014	&7.258939308	&9.869837922	&12.03298442	&13.85413531	&15.4082711	&16.7499948	&17.91999794	&18.94921795	&19.86159138\\ 4.147965319	&7.519876512	&10.31398664	&12.66663799	&14.67454391	&16.40815817	&17.91999794	&19.24999918	&20.42906542	&21.48148136\\ 4.229930538	&7.735489984	&10.68747581	&13.20708952	&15.38264829	&17.27999802	&18.94921795	&20.42906542	&21.74999987	&22.93628804 \end{bmatrix} $$       (5.2.exp.5)
 * style="width:10%; |
 * style="width:10%; |
 *

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

$$ \displaystyle F=\begin{bmatrix} -194.4238\\ 556.4552\\ -794.5674\\ 388.5998\\ 355.4029\\ -550.4317\\ 202.6984\\ 13.1421\\ 77.9578\\ -68.7552\\ -267.2934\\ 294.8667\\ 224.2897\\ -420.1257 150.8844 \end{bmatrix} $$       (5.2.exp.5) Solving for the general solution with n=6 gives an error of 2.83E-06 at u(.5) and a general solution of:
 * style="width:10%; |
 * style="width:10%; |
 *

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

$$ \displaystyle u^{h}=-167.2599*e^{1*(x-1)}	487.6591e^{2*(x-1)}	-808.7098e^{3*(x-1)}	770.2175e^{4*(x-1)}	-393.7902e^{5*(x-1)}	83.8821e^{6*(x-1)}+32.0012 $$       (5.2.exp.5) The boundary condition u(1)=4 is kept and the exact solution is very close to the trial solution:
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

1. Explain how the basis functions satisfy the constraint breaking solution (CBS).
The following are the basis functions:

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

$$\displaystyle \sum_{i=1}^m \prod_{k=1,i\ne k}^m \frac{(x-x_k)}{(x_i-x_k)} {m=2,4,6,8....}$$ (5.3.1)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

3. Solve the following equation using these basis functions and matlab.

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

$$\displaystyle \frac{\partial}{\partial x} \left[ a_2(x) \frac{\partial u}{\partial x} \right ] + f(x,t) = \bar{m} \frac{\partial ^s u}{\partial t ^s} (x,t)$$ (5.3.2)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

with boundary conditions:

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

$$ n(1) a_2(1) \frac{\partial u}{\partial x}(1)=12 \qquad \Rightarrow \qquad \frac{\partial u}{\partial x}(1)= \frac{12}{5}$$ (5.3.3)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

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

$$\displaystyle u(0)=4$$ (5.3.4)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

4. Make Plots of u^h vs u and u^h(.5)-u(.5) vs m

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

$$\displaystyle u_m^h$$ vs $$\displaystyle u$$ (5.3.5)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

and,

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

$$\displaystyle u_m^h(0.5)-u(0.5)$$ vs $$\displaystyle m$$ (5.3.6)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

Part 1
These basis functions have the property where at each node there is a single function equal to one and the rest are zero, as seen in figure on part 2. So, all that needs to be done is set the coefficient of the single nonzero basis function to zero to meet the constraint breaking solution requirement.

Part 2
Using the following Matlab code for m=2:

With the Following Plot:

For M=4:

With Plot:

For m=6:

With Plot:

For m=8:

With Plot for m=8:

To get a better picture of the properties of these 8 basis functions, the following is a plot where the basis functions are only evaluated at the nodes:

Part 3
Starting with the weak form of equation 5.3.3 (see derivation of weak form in problem 5.1):
 * {| style="width:100%" border="0"

$$ \int_0^1 \frac{dw}{dx}(2+3x) \frac{du}{dx} dx = 12w(1) + \int_0^1 w(x)5x dx \qquad \forall w \quad with \quad w(0)=0. $$     (5.1.8)
 *

 * }

Now approximate the solution by:
 * {| style="width:100%" border="0"

$$\displaystyle u\simeq u^h = \sum_{i=1}^md_i\prod_{k=1,k\neq i}^m \frac{(x-x_k)}{(x_i-x_k)}=\sum_{i=1}^md_i*b_i(x) $$     (5.3.7)
 *

 * }

and approximate the weighting function by:

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

$$\displaystyle w\simeq w^h = \sum_{i=1}^mc_i\prod_{k=1,k\neq i}^m \frac{(x-x_k)}{(x_i-x_k)}\sum_{i=1}^mc_i*b_i(x) $$     (5.3.8)
 *

 * }

Where bi(x) has been substituted in for the basis functions for ease of finding the solution.

Substituting Equation 5.3.7 and 5.3.8 into 5.1.8 and using linearity of the integral operator gives after some rearrangement:

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

$$\displaystyle \sum_{i=1}^m[\int_0^1(\frac{d}{dx}c_ib_i(x))(2+3x)(\frac{d}{dx}d_ib_i(x)-12c_ib_i(1)-\int_0^1c_ib_i(x)5xdx] $$     (5.3.9)
 *

 * }

Factoring out the constant ci:

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

$$\displaystyle \sum_{i=1}^mc_i[\int_0^1\frac{d}{dx}b_i(x)(2+3x)(\frac{d}{dx}d_ib_i(x)-12b_i(1)-\int_0^1b_i(x)5xdx] $$     (5.3.10)
 *

 * }

To solve for the unknown coefficients simultaneously, we will move to write 5.3.10 in tensor form.

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

$$\displaystyle \vec C=\begin{bmatrix} \vec C_E \\ \vec C_F\end{bmatrix}=\begin{bmatrix} 0 \\ \vec C_F\end{bmatrix} $$     (5.3.11)
 *

 * }

The zero in equation comes from satisfying the constraint breaking solution where the coefficient at the natural boundary condition for the only nonzero basis function coefficient on the weighting function is zero. Hence the subscript E in equation 5.3.11 for essential. Looking at the coefficients for uh:

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

$$\displaystyle \vec d=\begin{bmatrix} \vec d_E \\ \vec d_F\end{bmatrix}=\begin{bmatrix} g \\ \vec d_F\end{bmatrix} $$     (5.3.12)
 *

 * }

Likewise here, g is the value at the essential boundary condition. Since there is only one nonzero basis function here, the coefficient (since the basis function is one) must be equal to the essential boundary condition value. The subscript F can be thought of as free and will be determined.

Now defining the following

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

$$\displaystyle \tilde k(b_i,b_j)=\int_0^1(\frac{d}{dx}b_i)(2+3x)(\frac{d}{dx}b_j)dx $$     (5.3.13)
 *

 * }

This shows how the far left integral term of equation 5.3.10 can be written in components of a matrix. Additionally defining:

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

$$\displaystyle K_{EE}=[\tilde k(b_1,b_1)] $$     (5.3.14)
 *

 * }

Where subscript one signifies the first basis functions.
 * {| style="width:100%" border="0"

$$\displaystyle K_{EF}=[\tilde k(b_1,b_j)] $$     (5.3.15)
 *

 * }

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

$$\displaystyle K_{FE}=[\tilde k(b_i,b_1)] $$     (5.3.16)
 *

 * }

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

$$\displaystyle K_{FF}=[\tilde k(b_{remaining i},b_{remaining j})] $$     (5.3.17)
 *

 * }

Giving:

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

$$\displaystyle \mathbf K=\begin{bmatrix} \mathbf K_{EE} \mathbf K_{EF} \\ \mathbf K_{FE} \mathbf K_{FF}\end{bmatrix} $$     (5.3.18)
 *

 * }

Lastly, combining the constants into:
 * {| style="width:100%" border="0"

$$\displaystyle \mathbf F = [\tilde f_i] $$     (5.3.19)
 *

 * }

Where,

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

$$\displaystyle \tilde f_i=12b_i(1)+\int_0^1b_i(x)5xdx $$     (5.3.20)
 *

 * }

Putting it all together:

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

$$\displaystyle \vec C \cdot \mathbf K\mathbf d=\begin{bmatrix}0\\\vec C_F\end{bmatrix}\cdot \Big (\begin{bmatrix} \mathbf K_{EE} \mathbf K_{EF} \\ \mathbf K_{FE} \mathbf K_{FF}\end{bmatrix}\begin{bmatrix}\mathbf g\\\vec d_F\end{bmatrix}-\begin{bmatrix}\mathbf F_E\\ \mathbf F_F\end{bmatrix} \Big )=0 $$     (5.3.21) Now, our weighting function is arbitrary so what is in parenthesis corresponding to CF must equal zero. Writing this out explicitly for four basis functions gives the following:
 *

 * }

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

$$\displaystyle \begin{bmatrix}K_{11}K_{12}K_{13}K_{14}\\K_{21}K_{22}K_{23}K_{24}\\K_{31}K_{32}K_{33}K_{34}\\K_{41}K_{42}K_{43}K_{44}\end{bmatrix}\begin{bmatrix}d_1\\d_2\\d_3\\d_4\end{bmatrix}=\begin{bmatrix}f_1+r_1\\f_2\\f_3\\f_4\end{bmatrix} $$     (5.3.22)
 *

 * }

In order to solve four our unknowns d2-d4, in terms of our knowns, we rearrange to get the following:

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

$$\displaystyle \begin{bmatrix}K_{22}K_{23}K_{24}\\K_{32}K_{33}K_{34}\\K_{42}K_{43}K_{44}\end{bmatrix}\begin{bmatrix}d_2\\d_3\\d_4\end{bmatrix}=\begin{bmatrix}f_2-K_{21}d_1\\f_3-K_{31}d_1\\f_4-K_{41}d_1\end{bmatrix} $$     (5.3.23)
 *

 * }

This solution method was applied in Matlab to give the following:

m=2

m=4

m=6

m=8

Part 4
The following are the resulting plots from Part 3: m=2

m=4

m=6

m=8

This Matlab code was used to plot uh-u vs m:

Giving:

Problem Description
Solve G1D1M1.D/D1d, for$$\frac{d}((2 + 3x)\frac) + 5x = 0$$, using weak form with LIBF (m=4, 6, 8, ..), until convergence of $${u^h}(0.5)to{\rm O}({10^{ - 6}})$$.
 * 1) Explain how LIBF are used as CBS
 * 2) Plot all LIBF used
 * 3) Use Matlab quad or WolfAlpha to integrate
 * 4) Plot $${u^h}$$ vs $$u$$, $${u^h}(0.5) - u(0.5)$$ vs m

Solution
(1)
 * {| style="width:100%" border="0"

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

{L_{i,m}}({x_j}) = {\delta _{i,m}} = \left\{ {\begin{array}{ccccccccccccccc} {1,forj = i} \\ {0,forj \ne i} \end{array}} \right.

$$ So for series of LIBF, at the essential boundary condition, there will be only one term does not equal to zero, the other terms equal to zero. So the CBS has been satisfied.
 *
(Eq 5.4.1)
 * }

(2) Basis function:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{L_{i,m}}(x) = \prod\limits_{\begin{array}{ccccccccccccccc} {k = 1} \\ {k \ne i} \end{array}}^m {\frac}

$$     (Eq 5.4.2)
 *

 * }

Figs of the basis functions

m=4:

m=6:

m=8：

(3) For weak form
 * {| style="width:100%" border="0"

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

w(x) = \sum\limits_{i = 1}^m {{c_i} \cdot {L_{i,m}}}

$$     (Eq 5.4.3)
 *

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

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

u(x) = \sum\limits_{i = 1}^m {{b_i} \cdot {L_{i,m}}}

$$     (Eq 5.4.4) To boundary condition
 *

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

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

w(1) = 0

$$     (Eq 5.4.5)
 *

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

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

{\alpha _m} = 0

$$     (Eq 5.4.6)
 *

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

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

u(1) = 4

$$     (Eq 5.4.7)
 *

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

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

{b_m} = 4

$$     (Eq 5.4.8) The weak form is:
 *

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

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

\left. {w(2 + 3x)\frac} \right|_0^1 - \int\limits_0^1 {\frac(2 + 3x)\fracdx} + \int\limits_0^1 {w \cdot 5x}  = 0

$$     (Eq 5.4.9)
 *

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

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

- w(0)(2 + 3*0)\frac(0) - \int\limits_0^1 {\frac(2 + 3x)\fracdx} + \int\limits_0^1 {w \cdot 5x}  = 0

$$     (Eq 5.4.10)
 *

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

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

- w(0)*12 + \int\limits_0^1 {w \cdot 5x} = \int\limits_0^1 {\frac(2 + 3x)\fracdx}

$$     (Eq 5.4.11)
 *

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

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

\Rightarrow \sum\limits_i {{C_i}\left[ {\sum\limits_j  - {F_i}} \right]}  = 0

$$     (Eq 5.4.12) Where
 *

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

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

{K_{ij}} = \int\limits_0^1 {{L_{i,m}}'(2 + 3x){L_{j,m}}'}dx

$$     (Eq 5.4.13)
 *

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

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

{F_i} = \int\limits_0^1 {{L_{i,m}} \cdot 5x}dx + w(0)*12

$$     (Eq 5.4.14) In this static case
 *

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

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

Kd = Fwhere{F_F} - {K_{FE}}g

$$     (Eq 5.4.15) Solve for d
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{u^h} = \sum\limits_{i = 1}^m  \cdot {L_{i.m}}

$$     (Eq 5.4.16)
 *

 * }

Plot $${u^h}$$ vs $$u$$

m=4:

m=6:

m=8：

Calculate for error

Error vs m :

Problem 5.5.1 and 5.5.2
'''5.5.1 For the disc problem find node information : nodes numbers and their coordinates, Element numbers and their respective node numbers. 5.5.2 For the same disc problem generate 3 meshes with triangle elements by changing number of elements. ''' Eml5526.s11.team7.Srilalithkumar 16:03, 22 March 2011 (UTC)

Problem 5.5.3: CalculiX CrunchiX (CCX) for Dummies
Adopted from CCX User's Manual and Mechanical Hack's website. This walk through provides the basic overview of CCX. Experimentation and actual development of input files will be required on one's on.

Note: Previous installation of Calculix in Windows additionally installed CCX, so one can immediately begin to follow this example.

CCX is computational tool based on the information the user puts into the input file. CCX is constructed from FORTRAN77 and C programing languages, with a definite emphasis on FORTRAN. In relation to Calculix, CCX is considered the main subroutine of the total program Calculix.

CCX uses what it calls an "input deck" which specifies what will be the input. This input deck is made up of two parts, the model and step portions. Inside of these portions keywords are used. These keywords can be thought of as commands. These keywords belong either to the model or the step portion, and the model portion MUST be used before the step portion. To get a visual, look at the following figure built off of Mechanical Hack's website:

The *WORD are the keywords, or commands. The *HEADING command allows for data to be entered on the following line for the benefit of the user. The next set of commands describes the material that will be analyzed with CCX. This data can either be entered through *INCLUDE, which copies the data from the specified file (who must be in the designated path for CCX) to the input, or it can be entered explicitly as seen in the following code from the test folder on the Calculix website:

A search of the users manual provides additional methods of inputting data if needed. The remaining parts of this model section adds the necessary attributes relating to the physics of the model (boundary condition, physical properties, etc). This can be seen with the *BOUNDARY command, which allows for the proper formatting of the data being read from the input FIXED_123.BOU to be read as boundary conditions. Next the *MATERIAL is used to allow for assignment through NAME=EL of a name to the model being created. The keyword *ELASTIC allows for assignment of the elastic properties of the material. There are a multitude of additional keywords to completely specify the material of the problem. These are in the user's manual as well, and an example of them can be seen here:

Now that the material has been specified the keyword step can be used. Step allows for a quasi-static analysis. For example, in the picture a dynamic load is applied, so a step in time can be used to find the final results. End step signifies the end of the step, and there is a limit to the number of steps the program is capable of running.

With the basics of the input file created, the results are found by clicking on Tools, Calculate:

This splits the screen in two and shows the command portion of Calculix. If the input file was created correctly, the following "Job finished, Exit code:0" will be seen as well as the output.

Additionally output can be specified to a file for post processing with CGX (as seen in the figure above with the *NODE FILE and *EL FILE commands. If this happens, the data is sent to a .frb file which is post processed simply by Tools->Post Process, or Shift-F10.

Problem 5.6: Quadratic Lagrangian Element Basis Functions (QLEBF)
Plot three figures, similar to those in lecture 30, for the quadratic Lagrangian element basis functions. Consider the case where the distance between nodes is not equal.

Solution

Comments:
 * The first three quadratic basis functions are defined as:
 * $$\displaystyle b_1(x)=\frac{(x-x_2)(x-x_3)}{(x_1-x_2)(x_1-x_3)}$$,
 * $$\displaystyle b_2(x)=\frac{(x-x_1)(x-x_3)}{(x_2-x_1)(x_2-x_3)}$$,
 * $$\displaystyle b_3(x)=\frac{(x-x_1)(x-x_2)}{(x_3-x_1)(x_3-x_2)}$$.
 * where $$\displaystyle x_N$$ represents the x value of the global node 'N'. Subsequent basis functions (i.e. $$\displaystyle b_4, \ b_5, \ b_6, \ \text{etc.}$$) follow the same form.

 * Each element is not discretized uniformly. In other words each node is not equidistant from adjacent nodes. The x-position for each node was arbitrarily chosen and are given as:
 * $$\displaystyle \qquad x_1=1, \qquad x_2=1.75, \qquad x_3=3, \qquad x_4=3.5, \qquad \text{and} \qquad x_5=5$$.

 * The sum of all basis functions at any point along 'x' is one.
 * The first three plots of Figure 5.6 illustrate the global representation. The last plot of Figure 5.6 shows the case for the individual element $$\displaystyle \omega^{(1)}$$ and is defined between global nodes 1 and 3. A second element, say $$\displaystyle \omega^{(2)}$$, would then be defined between global nodes 3 and 5.
 * While each element contains three nodes, each node is assigned a "local" node number from 1 to 3. Specifically for the case of element 1 in Figure 5.6, the global node numbers are the same as the local node number. However, for element 2, global node 3 corresponds to local node 1, global node 4 corresponds to local node 2, and global node 5 corresponds to local node 3.
 * Since it can easily be confusing when correlating global node numbers and local node numbers, the relationship between the two can be generalized using the equation below.

 * $$\displaystyle I = i+(3-1)(e-1) \qquad \text{(for the case of 3 nodes per element)}$$
 * where $$\displaystyle I$$ is the global node number, $$\displaystyle i$$ is the local node number, and $$\displaystyle e$$ is the element number.

Problem Description
Solve the general 1-D model for an elastic bar with data set 1b (G1DM1.0/D1b) using the Lagrangian Interpolation Basis Function until convergence of $$\displaystyle u^h (0.5) \rightarrow O(10^{-6}) $$.
 * 1. Explain how the basis functions satisfy the constraint breaking solution (CBS).
 * 2.Plot these basis functions.
 * 3.Use matlab to integrate.
 * 4.Make Plots of u^h vs u and u^h(.5)-u(.5) vs m

Solution
The Lagrangian Element Basis Function (LEBF) is given by the following equation and generates a polynomial of degree $$\displaystyle n-1$$:

 * $$L_{i,n}\left (x\right )=\prod_{k=1}^{n}\frac{x-x_{k}}{x_{i}-x_{k}},k\neq i$$

For a linear function the number of nodes is 2 ($$\displaystyle n=2$$). For uniform discretization with $$\displaystyle nel=4$$ the domain $$\displaystyle \Omega$$ will be subdivided into 4 equal length elements each having two nodes. The LEBFs for the first element $$\displaystyle n=1$$ over the elemental domain $$\displaystyle \omega^{(1)}=[0,.25]$$ can be written as follows:

 * $$L_{1,2}^{(1)}\left (x\right)=b_{1}^{(1)}=\frac{(x-x_2)}{x_1-x_2}=\frac{(x-.25)}{0-.25}=-4(x-.25)$$

 * $$L_{2,2}^{(1)}\left (x\right)=b_{2}^{(1)}=\frac{(x-x_1)}{x_2-x_1}=\frac{(x-0)}{.25-0}=4(x)$$

Recall the constraint breaking solution (CBS) states the basis functions must satisfy $$\displaystyle \left\{ b_i(x), i=0,1,2,...\right\}$$ such that $$\displaystyle b_0(\beta) \neq 0$$ and $$\displaystyle b_i(\beta) = 0$$ for $$\displaystyle i=1,2,...,n$$.

It can be seen that these basis functions for the first element satisfy the CBS at each node. For the first node, the basis function has a value of one at that node and a value of zero at every other node (there is only one other node for the linear LEBF). In other words the basis function is equal to the kronicker delta at each node.

 * $$b_{1}^{(1)}(\beta^{(1)}_1)=L_{1,2}^{(1)}\left (0\right)=b_{1}^{(1)}(0)=-4(0-.25)=1$$

 * $$L_{1,2}^{(1)}\left (.25\right)=b_{1}^{(1)}(.25)=-4(.25-.25)=0$$

Similarly for the second node, the basis function has a value of one at that second node and a value of zero at the first node.

A Matlab script was written to plot the LEBFs at each node for each element. The Matlab script is given followed by the plots. The first plot shows each of the LEBFs plotted across the entire domain. The second plot is more informative as the plot is divided into the elemental domains and lines corresponding to the values zero and one are present. The legend states the node number, $$n$$, and element number, $$e$$, of each LEBF.

As with the CBS, the coefficient at the essential boundary condition is equal to the essential boundary condition because the basis function is equal to one at the essential boundary condition:

 * $$g = \mathbf{d_{1}^{(1)}}\mathbf{b_{1}^{(1)}}(0)=\mathbf{d_{1}^{(1)}}*1=\mathbf{d_{1}^{(1)}}=4$$

The element matrices are formed as in problems 5.1 and 5.2 as follows:

 * $$\displaystyle \mathbf{F}^{e}_{F}=b_{i}^{(1)}(\alpha_{i}^{e} )h + \int_{\omega^{e}}^{.} {{b_{i}^{(1)}}fdx}$$

\displaystyle

h = {a_2}(\alpha )\frac = (2+3\alpha)\frac = 5*12/5=12

$$

 * $$\displaystyle \mathbf{F_{F}^{(1)}}=\mathbf{F}^{(1)}_{2}=b_{2}^{(1)}(.25)h + \int_{0}^{.25} {{b_{2}^{(1)}}fdx}=1*12+\int_{0}^{.25} 4x(5x)=12.104167$$

$$ \displaystyle

\therefore {F_F} = \left[ {\begin{array}{ccccccccccccccc} {12.104167} \end{array}} \right]

$$

 * $$\displaystyle \mathbf{K}^{e}_{ij}=\int_{\omega^{e}}^{.} b_{i}^{e'}(x^e)a_2(x^e)b_{j}^{e'}(x^e)dx$$

\displaystyle

\mathbf{K^{(1)}_{FE}} = \left\{ {\mathbf{K^{(1)}_{1j}};j = 2,...,n} \right\}

$$

\displaystyle

\mathbf{K^{(1)}_{12}}= \int\limits_0^{.25} b_{1}^{(1)'}a_2b_{2}^{(1)'}=\int\limits_0^{.25}-4*(2+3x)*4=-32(.25)-24(.25)^2=-9.5

$$

\displaystyle \therefore \mathbf{K^{(1)}_{FF}} = \begin{bmatrix} -9.5 \end{bmatrix} $$

\displaystyle

\mathbf{K^{(1)}_{FF}} = \left\{ {\mathbf{K^{(1)}_{ij}};i,j = 2,...,n} \right\}

$$

\displaystyle

\mathbf{K^{(1)}_{22}}= \int\limits_0^{.25} b_{2}^{(1)'}a_2b_{2}^{(1)'}=\int\limits_0^{.25}4*(2+3x)*4=32(.25)+24(.25)^2=9.5

$$

\displaystyle \therefore \mathbf{K^{(1)}_{FF}} = \begin{bmatrix} 9.5 \end{bmatrix} $$

 * $$\mathbf{K^{(1)}} \mathbf{d^{(1)}} = \mathbf{F_{F}^{(1)}} - \mathbf{K_{FE}^{(1)}} g$$

 * $$\mathbf{d_{2}^{(1)}} = \mathbf{K}_{FF}^{(1)-1}( \mathbf{F_{F}^{(1)}} - \mathbf{K_{FE}^{(1)}} g)$$

 * $$\therefore \mathbf{d_{2}^{(1)}} = 5.5208$$

Now the process is repeated for each element. The matrices are assembled into global matrices and all of the coefficients can be determined. The approximate solution is formed for each element by taking the coefficients times the basis functions. The approximate solutions for each element make up the global approximate solution.

 * $$\displaystyle \mathbf{K}_{ij} = \sum_{e=1}^{nel}\mathbf{K}^{e}_{ij}$$

 * $$\displaystyle \mathbf{F}_{i} = \sum_{e=1}^{nel}\mathbf{F}^{e}_{i}$$

 * $$\displaystyle u^{he}(x)= \sum_{j=1}^{n}d_jb_j(x),\quad \quad u^{h}(x)=\sum_{j=1}^{nel}u^{he}(x) $$

A Matlab script was written to generate the plots. The results and script are given below:

Problem Description
Solve the general 1-D model for an elastic bar with data set 1 (G1DM1.0/D1) using the Linear Lagrangian element basis function (LLEBF) with uniform discretization(equidistant element nodes, nel=4, 6, 8).
 * 1.Explain how the basis functions satisfy the constraint breaking solution (CBS).
 * 2.Plot these basis functions.
 * 3.Use matlab to integrate.
 * 4.Make Plots of u^h vs u and u^h(.5)-u(.5) vs m

Explain how LLEBF are used as CBS
This problem is similar to problem 5.7 except switch the location of natural and essential boundary conditions.

The Lagrangian Element Basis Function (LEBF) is given by the following equation and generates a polynomial of degree $$\displaystyle n-1$$:

 * $$L_{i,n}\left (x\right )=\prod_{k=1}^{n}\frac{x-x_{k}}{x_{i}-x_{k}},k\neq i$$

For a linear function the number of nodes is 2 ($$\displaystyle n=2$$). For uniform discretization with $$\displaystyle nel=4$$ the domain $$\displaystyle \Omega$$ will be subdivided into 4 equal length elements each having two nodes. The LEBFs for the first element $$\displaystyle n=1$$ over the elemental domain $$\displaystyle \omega^{(1)}=[0,.25]$$ can be written as follows:

 * $$L_{1,2}^{(1)}\left (x\right)=b_{1}^{(1)}=\frac{(x-x_2)}{x_1-x_2}=\frac{(x-.25)}{0-.25}=-4(x-.25)$$

 * $$L_{2,2}^{(1)}\left (x\right)=b_{2}^{(1)}=\frac{(x-x_1)}{x_2-x_1}=\frac{(x-0)}{.25-0}=4(x)$$

Recall the constraint breaking solution (CBS) states the basis functions must satisfy $$\displaystyle \left\{ b_i(x), i=0,1,2,...\right\}$$ such that $$\displaystyle b_0(\beta) \neq 0$$ and $$\displaystyle b_i(\beta) = 0$$ for $$\displaystyle i=1,2,...,n$$.

It can be seen that these basis functions for the first element satisfy the CBS at each node. For the first node, the basis function has a value of one at that node and a value of zero at every other node (there is only one other node for the linear LEBF). In other words the basis function is equal to the kronicker delta at each node.

 * $$b_{1}^{(1)}(\beta^{(1)}_1)=L_{1,2}^{(1)}\left (0\right)=b_{1}^{(1)}(0)=-4(0-.25)=1$$

 * $$L_{1,2}^{(1)}\left (.25\right)=b_{1}^{(1)}(.25)=-4(.25-.25)=0$$

Similarly for the second node, the basis function has a value of one at that second node and a value of zero at the first node.

Therefore, LLEBF satifies CBS

title: User:Eml5526.s11.team7/HW5


text: File:Nm1 s11 team1 HW5 p4 fig a.png

title: File:Nm1 s11 team1 HW5 p4 fig a.png


text: File:Nm1 s11 team1 HW5 p4 fig b.png

title: File:Nm1 s11 team1 HW5 p4 fig b.png