text:
Instructional design/Adult learning/Choosing Methods for a Presentation

Choosing Instructional Methods
When putting together an instructional presentation For example, if participants need to learn how to complete a form - give them the form to complete, don’t give them a lecture! If you want participants to solve a problem – give them a scenario, let them work it out in a group, or have them role play the situation.
 * take into consideration the characteristics of adult learners,
 * choose instructional methods that are appealing and effective with adult learners, and
 * choose instructional methods that match the content you want to teach.

Case Examples
Here are two examples of presentations and how the instructor matched instructional methods with adult learner characteristics and instructional content:

Worker Safety
Mike is a county sheriff who has been asked to give a two hour presentation on basic safety and self-defense for case managers in a social service organization. Case managers in this organization go into the homes of families and work with them on a variety of issues including addiction, parenting skills, and mental health management. Mike wants to demonstrate some of the physical signs workers may see in a client that are cues of escalation that may lead to violence. Mike also wants to demonstrate some simple self-defense moves a worker can use if physically attacked. Mike would also like to have participants role play an attack with him so they can practice using the moves. In order to keep the presentation realistic and relevant Mike wants to use several Case Scenarios based on real cases he was involved with. He plans to describe the case and ask the participants questions about what they would do in that situation. Mike knows that this is an important topic to participants and so he makes sure to leave time at the closing for a group discussion directed by the participants regarding any of the materials in his presentation.

Content:
 * Safety and self-defense strategies

Adult Learner Characteristics taken into account:
 * Self-concept
 * Readiness to learn
 * Orientation to learn
 * Motivation to learn

Instructional Methods used:
 * Demonstration
 * Role play
 * Case scenarios
 * Group discussion

Court Documents
Carey is an attorney that has been asked to give a two hour presentation to child protective service (CPS) workers on the content entered in two common legal documents that are submitted to the court. The content in the court documents is entered into a piece of software by answering screen prompts; the software then generates the completed form. One challenge Carey faces is that she will not have access to the software during the training.

Carey knows that CPS workers are frequently nervous about completing court documents and this training will help prepare them, so she knows her participants will be ready to learn. Carey knows that adults want to see how the information is directly connected to how they can use it so Carey includes screen shots of the software in the materials and uses real cases to demonstrate what content to include. Carey knows that adults want to direct their own learning so she plans to present the materials in small sections and allow participants time to ask questions and direct where the learning will take them. Carey knows that the participants will likely be motivated to learn if the information she provides can help them with the cases they are working on. Before the training Carey collects information on some common themes of CPS cases to use as case studies for the participants to work with as a team project.

Content:
 * Court documents

Adult Learner Characteristics taken into account:
 * Self-concept
 * Experience
 * Readiness to learn
 * Orientation to learn
 * Motivation to learn

Instructional Methods used:
 * Demonstration
 * Case studies
 * Team project
 * Group discussion

Practice Activity
Now it’s time to practice your skills. Imagine you have been asked to teach the courses below. Read the course descriptions and choose instructional methods to present the content that address adult learner characteristics and match the type of content. Identify the adult learner characteristics your presentation takes into account and identify the instructional methods used. Write your ideas down on a piece of paper. When you are finished click on the corresponding link to show the ideas of an expert.

Beginning Quilting
This class is an exciting introduction to quilting for new quilters! In this two hour class you will learn basic quilting skills such as how to read a quilt block pattern, how to cut quilt block pieces out with a rotary cutter, how to sew a quilter’s quarter-inch seam, and how to construct a simple quilt block. You will have several simple projects to choose from to learn these skills. You will need to bring your own sewing machine, rotary cutter, quilter’s ruler, and five fat quarters of coordinating fabrics from your stash.

Flora Photography
This class will give photo enthusiasts the opportunity to photograph still shots of flora in the county park conservatory and formal garden while receiving instruction from a professional photographer. This course will include access to the conservatory, a group lecture, and both one-on-one and group instruction. This program is designed for people with some basic cameral knowledge. Participants will need to bring a digital and/or a 35mm camera.

Next Steps
Now that you have had an opportunity to practice recognizing and addressing adult learner characteristics through choosing appropriate instructional methods you are ready to start putting things together. The next step in the lesson will get you on your way.

Click Wrap up and Extension to continue.


text:
Instructional design/Adult learning/Wrap up and Extension

Wrap Up
As a subject matter experts, when you create a lesson, you hope that the instructional event will be relevant, practical, engaging, supportive, and a socially interactive learning experience for adult learners. By remembering and utilizing the techniques provided in this session you’ll be closer to providing this experience for your learner each and every time. When creating instructional events, remember to allow the following facts and characteristics regarding adult learners to guide your decisions:

 * 1)	Take into consideration the characteristics of adult learners
 * a.	Independent and self-directed
 * i.	Self-concept
 * b.	Have a wealth of life experiences to draw on
 * i.	Experience
 * c.	Motivated to learn when they can see the connection to how their life is affected by the instruction.
 * i.	Readiness to learn
 * ii.	Orientation to learn
 * iii.	Motivation to learn

 * 2)	 Choose instructional methods that are appealing and effective with adult learners
 * a.	Project/Team Project
 * b.	Game
 * c.	Role Play
 * d.	Demonstration
 * e.	Case Study
 * f.	Group Discussion
 * g.	Lecture

 * 3)	 Select instructional methods that match the content you want to teach.

Extension
Now that you have become familiar with the various techniques for addressing the needs of adult learners, go do it! Feel free to come back and tell us how it went, and suggest possible pointers that your peers and fellow subject matter experts may find helpful as well. Good luck!

Click Resources to continue.

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


text:
Instructional design/Adult learning/Resources

Resources: Choosing Methods for Adult Learners
To read more about Adult Learning Theory:

 * Malcolm Knowles, informal adult education, self-direction and andragogy
 * Malcolm Knowles: Apostle of Andragogy
 * Andragogy
 * Andragogy + Pedagogy

To read more about developing subject matter content into effective lessons: Articulate.com

To read more about Instructional Methods: Olson, Dorsey, & Reigeluth - as found on page 22 of Instructional -Design Theories and Models

Click Instructional Design: Homepage to view other instructional design lessons.


title: File:Hw5.4-u=0.jpg


title: File:HW5Prob5.jpg


text:
User talk:Robert Flogaus-Faust

 Hello Robert Flogaus-Faust, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.
 * Explore Wikiversity with the links to your left.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.
 * Follow Wikiversity on twitter (http://twitter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! SB_Johnny talk 23:46, 22 March 2011 (UTC)

title: User talk:Robert Flogaus-Faust


title: File:HW5Prob5Romberg.jpg


text: File:Eml5526.s11.team4.simko.hw5.problem3.fig1.png


text: File:Nm1.s11.team4.HW5 P3.fig1.jpg

title: File:Nm1.s11.team4.HW5 P3.fig1.jpg


text: File:Nm1.s11.team4.HW5 P3.fig2.jpg

title: File:Nm1.s11.team4.HW5 P3.fig2.jpg


text:
User:Eml5526.s11.team3.akj/Homework 5

=Problem 5.1 Using appropriate basis functions (Polynomial, Fourier and Exponential basis functions) to obtain the trial solution from Weak Form=

Problem Statement
$$ \frac{\partial}{\partial x} \left[ (2+3x) \frac{\partial u}{\partial x} \right] + 5x = 0 $$, $$ \forall x \in \left[ 0,1 \right] $$, $$ u(0)=4 $$, $$ n(1)a_2(1)\frac{\partial u(x=1)}{\partial x}=12$$

$$\implies (1)(2+3)\frac{\partial u(x=1)}{\partial x}=12$$

$$\implies \frac{\partial u(x=1)}{\partial x}=\frac{12}{5}

$$

$$ \Gamma_g = [0] $$, $$ \Gamma_h = [1] $$

Solve the above problem using Weight Function with appropriate basis function (Polynomial basis function, Fourier basis function and Exponential basis function) until convergence of $$u^h(0.5)$$ to $$10^{-6}$$.

Plot $$ u, u^h $$, convergence (error vs. n)

Solution
The strong form of $$\text{G1DM1}.0/\text{D1}$$ is:
 * {| style="width:100%" border="0"

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

\begin{align} & \frac{d}{dx}[(2+3x)\frac{du}{dx}]+5x=0,_ – ^ – \forall x\in (0,1) \\ & u(x=0)=4,_ – ^ – \frac{du}{dx}(x=1)=\frac{12}{5} \\ \end{align}

$$ We can derive the weak form as:
 * }
 * {| style="width:100%" border="0"

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

\begin{align} & \int\limits_{0}^{1}{\frac{\partial w}{\partial x}(2+3x)\frac{\partial u}{\partial x}}dx-\int\limits_{0}^{1}{5xwdx}-(\frac{24}{5}+\frac{36}{5}x)w\left| _{x=1} \right.=0,_ – ^ – \forall w(x=0)=0\\ & u(x=0)=4 \\ \end{align}

$$ The discretized weak form is
 * }
 * {| 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} & \underbrace{[\int\limits_{0}^{1}{\frac{\partial {{b}_{i}}(x)}{\partial x}(2+3x)\frac{\partial {{b}_{j}}(x)}{\partial x}}dx]}_{{d}_{i}}=\underbrace{[\int\limits_{0}^{1}{5x{{b}_{i}}(x)dx}+(\frac{24}{5}+\frac{36}{5}x){{b}_{i}}(x)\left| _{x=0} \right.]}_ \\ & {{d}_{0}}=4 \\ & i=1,2,\cdots \\ & j=1,2,\cdots \\ \end{align}

$$ The appropriate Polynomial basis function is
 * }
 * {| style="width:100%" border="0"

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

{{f}_{p}}=\{{{x}^{j}},j=0,1,2,\cdots \}

$$ The appropriate Fourier basis function is
 * }
 * {| style="width:100%" border="0"

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

{{f}_{F}}=\{\cos (jx)-1,\sin (kx),j=0,1,2,\cdots ,k=1,2,\cdots \}

$$ The appropriate Exponential basis function is
 * }
 * {| style="width:100%" border="0"

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

{{f}_{e}}=\{{{e}^{jx}}-1,j=0,1,2,\cdots \}

$$ The exact solution of governing equation is
 * }
 * {| style="width:100%" border="0"

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

u(x)=\frac{241}{54}\ln (\frac{2+3x}{2})+\frac{1}{36}(20x-15{{x}^{2}}+144)

$$
 * }

Using Polynomial basis function
The trial solution is
 * {| style="width:100%" border="0"

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

u_{p}^{h}(x)={{d}_{0}}{{x}^{0}}+{{d}_{1}}{{x}^{1}}+{{d}_{2}}{{x}^{2}}+\cdots +{{d}_{j}}{{x}^{j}},_ – ^ – j=0,1,2,\cdots

$$ Since the trial solution must satisfy the essential boundary condition,$$u(x=1)=4$$, substituting $$x=0$$into the trial solution, we have $${{d}_{0}}=4$$. In order to solve for the rest coefficient$${{d}_{j}},_ – ^ – j=1,2,\cdots $$, we construct the stiffness matrix and force matrix as follow:
 * }
 * {| style="width:100%" border="0"

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

{{K}_{F\times F}}=[{{k}_{ij}}]=[\int\limits_{0}^{1}{i{{x}^{i-1}}(2+3x)j{{x}^{j-1}}}dx]

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

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

{{F}_{F\times 1}}=[{{f}_{i\times 1}}]=[\int\limits_{0}^{1}{5x{{x}^{i}}}+(\frac{24}{5}+\frac{36}{5}*1){{1}^{i}}]

$$ Using Matlab to construct above two matrices until satisfying the convergence criterion, we finally 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" |

{{K}_{FF}}=\left( \begin{matrix} 7/2 & 4 & 17/4 & 22/5 & 9/2 & 32/7 & 37/8 & 14/3\\ 4 & 17/3 & 33/5 & 36/5 & 160/21 & 111/14 & 49/6 & 376/45\\ 17/4 & 33/5 & 81/10 & 64/7 & 555/56 & 21/2 & 329/30 & 624/55\\ 22/5 & 36/5 & 64/7 & 74/7 & 35/3 & 188/15 & 728/55 & 152/11\\ 9/2 & 160/21 & 555/56 & 35/3 & 235/18 & 156/11 & 665/44 & 620/39\\ 32/7 & 111/14 & 21/2 & 188/15 & 156/11 & 171/11 & 217/13 & 1608/91\\ 37/8 & 49/6 & 329/30 & 728/55 & 665/44 & 217/13 & 469/26 & 96/5\\ 14/3 & 376/45 & 624/55 & 152/11 & 620/39 & 1608/91 & 96/5 & 308/15\\ \end{matrix} \right)

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

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

{{F}_{F\times 1}}={\left( \begin{matrix} 41/3 \\ 53/4 \\ 13 \\ 77/6 \\ 89/7 \\ 101/8 \\ 113/9 \\ 25/2\\ \end{matrix} \right)}

$$ By the equation that$$Kd=F$$, we can solve the rest of $$d$$ as
 * }
 * {| style="width:100%" border="0"

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

{{d}_{F\times 1}}={\left( \begin{matrix} -0.23479307 \\ 1.2522298 \\ -3.0192656 \\ 4.5191594 \\ -5.0302374 \\ 4.9208409 \\ -5.4294546 \\ 7.2497811\\ \end{matrix} \right)}

$$ Hence, the trial solution of the governing equation is
 * }
 * {| style="width:100%" border="0"

$$\displaystyle \begin{align} & U_{F}^{h}(x)= - 0.23479307\;x^8 + 1.2522298\;x^7 - 3.0192656\;x^6 + 4.5191594\;x^5\\ & - 5.0302374\;x^4 + 4.9208409\;x^3 - 5.4294546\;x^2 + 7.2497811\;x + 4\\ \end{align}
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

$$
 * }

Using Fourier basis function
The trial solution is
 * {| style="width:100%" border="0"

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

\begin{align} & u_{F}^{h}(x)={{d}_{0}}[\cos [0*(x)]-1]+{{d}_{1}}[\cos [1*(x)]-1]+{{d}_{2}}\sin [1*(x)]+\cdots \\ & \begin{matrix} {} & {} \\ \end{matrix}+{{d}_{j-1}}[\cos [\frac{j}{2}(x)]-1]+{{d}_{j}}\sin [\frac{j}{2}(x)],j=0,1,2,\cdots  \\ \end{align}

$$ Since the trial solution must satisfy the essential boundary condition,$$u(x=1)=4$$, substituting $$x=1$$into the trial solution, we have $${{d}_{0}}=4$$. In order to solve for the rest coefficient$${{d}_{j}},_ – ^ – j=1,2,\cdots $$, we construct the stiffness matrix and force matrix as follow:
 * }
 * {| style="width:100%" border="0"

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

{{K}_{F\times F}}=[{{k}_{ij}}]=[\left\{ \begin{matrix} \int\limits_{0}^{1}{i\cos i(x)(2+3x)j\cos j(x)}dx,_ – ^ – when_ – ^ – i,j_ – ^ – both_ – ^ – are_ – ^ – even_ – ^ – number \\ \int\limits_{0}^{1}{i\sin i(x)(2+3x)j\sin j(x)}dx,_ – ^ – when_ – ^ – i,j_ – ^ – both_ – ^ – are_ – ^ – odd_ – ^ – number \\ \int\limits_{0}^{1}{i\cos i(x)(2+3x)j\sin j(x)}dx,_ – ^ – others \\ \end{matrix} \right.]

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

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

{{F}_{F\times 1}}=[{{f}_{i\times 1}}]=[\left\{ \begin{matrix} \int\limits_{0}^{1}{5x\sin i(x)}dx-(12+18*0)\sin i(1),_ – ^ – when_ – ^ – i_ – ^ – is_ – ^ – even_ – ^ – number \\ \int\limits_{0}^{1}{5x[\cos i(x)-1]}dx-(12+18*0)[\cos i(1)-1],_ – ^ – when_ – ^ – i_ – ^ – is_ – ^ – odd_ – ^ – number \\ \end{matrix} \right.]

$$ Using Matlab to construct above two matrices until satisfying the convergence criterion, we finally 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" |

{{K}_{FF}}=\left( \begin{matrix} 1.1444 & -1.3612  &  3.2564 &  -0.5408  &  3.7008  &  2.3208  &  1.2335 &   4.7249\\ -1.3612 &   2.3556 &  -4.1866  &  2.4001 &  -5.8462 &  -0.0674 &  -4.7299 &  -2.9460\\ 3.2564 &  -4.1866  &  9.5121 &  -2.3503 &  11.6194  &  5.4649  &  5.7892  & 12.8497\\ -0.5408 &   2.4001  & -2.3503  &  4.4879 &  -5.4725 &   5.3501 &  -8.9290 &   3.8999\\ 3.7008 &  -5.8462 &  11.6194 &  -5.4725  & 16.8127 &   2.2054 &  14.2443 &  12.2128\\ 2.3208 &  -0.0674  &  5.4649  &  5.3501  &  2.2054 &  14.6873 &  -9.6620 &  19.6948\\ 1.2335 &  -4.7299  &  5.7892  & -8.9290 &  14.2443  & -9.6620 &  23.4828 &  -3.0985\\ 4.7249 &  -2.9460 &  12.8497  &  3.8999 &  12.2128 &  19.6948 &  -3.0985 &  32.5172\\ \end{matrix} \right)

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

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

{{F}_{F\times 1}}={\left( \begin{matrix}  -6.1075 \\ 11.6035 \\ -18.9907 \\  13.0886 \\ -27.2503 \\  3.4218 \\ -23.8065 \\ -8.5011\\ \end{matrix} \right)}

$$ By the equation that$$Kd=F$$, we can solve the rest of $$d$$ as
 * }
 * {| style="width:100%" border="0"

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

{{d}_{F\times 1}}={\left( \begin{matrix} 0.90619376 \\ -4.1185023 \\ -3.4991531 \\ -30.621018 \\ -7.205183 \\ -0.42479515 \\ -47.408642 \\ 48.575238 \\ \end{matrix} \right)}

$$ Hence, the trial solution of the governing equation is
 * }
 * {| 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} & U_{F}^{h}(x)=\text{0}.\text{9}0\text{619376cos}\left( \text{4x} \right)-\text{4}.\text{1185023cos}\left( \text{3x} \right)-\text{3}.\text{4991531cos}\left( \text{2x}\right) \\ & -\text{30}.\text{621018sin}\left( \text{2x} \right)-\text{7}.\text{205183sin}\left( \text{3x}\right)-\text{0}.\text{42479515sin}\left( \text{4x}\right) \\ & -\text{47}.\text{408642cos}\left( \text{x}\right)+\text{48}.\text{575238sin}\left( \text{x}\right)+\text{36}.\text{69718}\\ \end{align}

$$
 * }

Using Exponential basis function
The trial solution is
 * {| style="width:100%" border="0"

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

u_{e}^{h}(x)={{d}_{0}}({{e}^{0*(x)}}-1)+{{d}_{1}}({{e}^{1*(x)}}-1)+{{d}_{2}}({{e}^{2*(x)}}-1)+\cdots +{{d}_{j}}({{e}^{j*(x)}}-1),_ – ^ – j=0,1,2,\cdots

$$ Since the trial solution must satisfy the essential boundary condition,$$u(x=1)=4$$, substituting $$x=1$$into the trial solution, we have $${{d}_{0}}=4$$. In order to solve for the rest coefficient$${{d}_{j}},_ – ^ – j=1,2,\cdots $$, we construct the stiffness matrix and force matrix as follow:
 * }
 * {| style="width:100%" border="0"

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

{{K}_{F\times F}}=[{{k}_{ij}}]=[\int\limits_{0}^{1}{i{{e}^{i(x)}}(2+3x)j{{e}^{j(x)}}}dx]

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

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

{{F}_{F\times 1}}=[{{f}_{i\times 1}}]=[\int\limits_{0}^{1}{5x[{{e}^{i(x)}}-1]dx}-(12+18*0)[{{e}^{i(1)}}-1]]

$$
 * }

Using Matlab, if we make the convergence criterion a little larger, that’s to $${{10}^{-5}}$$, we can get following results.

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

$$\displaystyle {{K}_{FF}}=1.0e+009 *\left( \begin{matrix} 0.0000 &  0.0000  &  0.0000  &  0.0000  &  0.0000  &  0.0000  &  0.0000  &  0.0000  &  0.0001 \\ 0.0000  &  0.0000  &  0.0000  &  0.0000  &  0.0000  &  0.0000  &  0.0001  &  0.0002  &  0.0005 \\ 0.0000  &  0.0000  &  0.0000  &  0.0000  &  0.0000  &  0.0001  &  0.0002  &  0.0006  &  0.0017 \\ 0.0000  &  0.0000  &  0.0000  &  0.0000  &  0.0001  &  0.0002  &  0.0007  &  0.0021  &  0.0058 \\ 0.0000  &  0.0000  &  0.0000  &  0.0001  &  0.0003  &  0.0008  &  0.0023  &  0.0065  &  0.0185 \\ 0.0000  &  0.0000  &  0.0001  &  0.0002  &  0.0008  &  0.0023  &  0.0068  &  0.0197  &  0.0565 \\ 0.0000  &  0.0001  &  0.0002  &  0.0007  &  0.0023  &  0.0068  &  0.0201  &  0.0586  &  0.1684 \\ 0.0000  &  0.0002  &  0.0006  &  0.0021  &  0.0065  &  0.0197  &  0.0586  &  0.1711  &  0.4935 \\ 0.0001  &  0.0005  &  0.0017  &  0.0058  &  0.0185  &  0.0565  &  0.1684  &  0.4935  &  1.4281 \\
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

\end{matrix} \right) $$
 * }

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

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

{{F}_{F\times 1}}=1.0e+005 *

$$ By the equation that$$Kd=F$$, we can solve the rest of $$d$$ as Hence, the trial solution of the governing equation is
 * }
 * {| 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} &u_{e}^{h}(x) = 507.7233\;e^{3x} - 455.35367\;e^{2x} - 371.33188\;e^{4x}+ 182.56105\;e^{5x} \\ & - 59.974767\;e^{6x} + 12.652506\;e^{7x} - 1.5521958\;e^{8x}+ 0.084261502\;e^{9x} \\ & + 250.24399\;e^{x} - 61.0526\\ \end{align}

$$
 * }

Comment
When we use the exponential basis function($$e^{(i(x))}-1$$) to obtain the trial solution from discretized weak form, the error at point x=0.5 never converges to $$10^{-6}$$. But if we change the accuracy to $$10^{-5}$$, it will converges to this far very fast.Among the two criterion for convergence, the continuity is satisfied, while the completeness is not. It's possible because that the exponential basis function doesn't contain a linear term or it can't used to approximate a linear function. So it's incomplete. There is also another problem rises from this assumption. Since there is certain equivalent relationship between exponential basis function and Fourier basis function, the same may also happen for Fourier basis function. But the Fourier basis function does converge to the accuracy of $$10^{-6}$$ here. Since it's much expensive to run the code for Fourier basis function, especially to a much accurate convergence, we don't have the chance to testify this inference.

title: User:Eml5526.s11.team3.akj/Homework 5


text: File:Fe1.s11.team1.hw5.node.element.info.pdf


text:
User:Egm6341.s11.team2.cleveland/HW5HW5

=Problem 5.7: =

Problem Statement:
Discuss the pros and cons of the following quadrature methods:

 * 1) Taylor's Series
 * 2) Composite Trapezoidal Rule
 * 3) Composite Simpson's Rule
 * 4) Romberg Table (Including Richardson's Extrapolation)
 * 5) Corrected Trapezoidal Rule

Pros:

 * 1) Local behavior about the point of expansion is very accurate, fast and computationally tractable.
 * 2) The number of operations are typically small for higher order approximation of simple functions, such as polynomials.
 * 3) With the aid of symbolic packages complex function behavior is easily found and accurate.

Cons:

 * 1) One needs to know the "smoothness" and global behavior of the function since computing derivative of an unbounded solution would be pointless.
 * 2) The number of operations may be large for a complex functions. For example, functions containing non-distributable products, quotients or composite trigonometric and exponential, to name a few.
 * 3) In addition, oscillatory functions, especially those having a small period(high frequency) are more cumbersome to approximate using a Taylor's Series.

Pros:

 * 1) Code implementation is very easy.
 * 2) Convergence for periodic functions is very quick when choosing to integrate over integer multiple of the period.
 * 3) The function (in the integrand) need only be twice continuously differentiable on the domain.

Cons:

 * 1) Steep concavity (or convexity) of the integrand can result in under or overestimates due to the second order continuously differentiable criteria. This means for a concave up integrand, and thus a positive second derivative, one will observe negative error and therefore an overestimate. Similarly, for a concave down (convex) integrand, and thus a negative second derivative, one will observe positive error and therefor an underestimate.
 * 2) Add more here!!!!!!!!!!!!!!!!!!!!

title: User:Egm6341.s11.team2.cleveland/HW5HW5


title: File:Fe1.s11.team1.hw5.fig5.1.jpg


title: File:Fe1.s11.team1.hw5.fig5.2.jpg


title: File:Fe1.s11.team1.hw5.fig5.3.jpg


text:
User:EML5526.S11.Team3.vnarayanan/Homework5

Problem Statement
Plot the Lagrangian Element basis function in case of a Quadratic element for the individual node and a combined plot.

Solution
A quadratic element has 3 nodes. Hence,

$$\;\;n = 3\;\;$$

For convenience we take

$$x_{1} = -1\; ; x_{2} = 0\; ;x_{3} = 1$$

The Lagrangian Interpolation Basis Function is given by,
 * {| style="width:100%" border="0"

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

L_{i,n}(x) = \prod_{k = 1}^{m}\frac{x-x_{k}}{x_{i}-x_{k}}\;\; k\neq i

$$
 * }

$$L_{1,3}(x) = \frac{(x-x_{2})(x-x_{3})}{(x_{1}-x_{2})(x_{1}-x_{3})} = \frac{(x-0)(x-1)}{(-1-0)(-1-1)} = \frac{x(x-1)}{2} = \frac{x^2-x}{2}$$

$$L_{2,3}(x) = \frac{(x-x_{1})(x-x_{3})}{(x_{2}-x_{1})(x_{2}-x_{3})} = \frac{(x-(-1))(x-1)}{(0+1)(0-1)} = \frac{(x+1)(x-1)}{-1} = 1-x^2$$

$$L_{3,3}(x) = \frac{(x-x_{1})(x-x_{2})}{(x_{3}-x_{1})(x_{3}-x_{2})} = \frac{(x-(-1))(x-0)}{(1+1)(1-0)}  = \frac{(x+1)(x)}{2} = \frac{x^2+x}{2}$$

Using the above expressions, the plots of the basis functions were developed using Matlab


text: File:FE1 Team6 Figure- 5.2.1.jpg

title: File:FE1 Team6 Figure- 5.2.1.jpg


text: File:FE1 Team6 Figure- 5.2.2.jpg

title: File:FE1 Team6 Figure- 5.2.2.jpg


text: File:FE1 Team6 Figure-5.7.1 LLEBF.png

title: File:FE1 Team6 Figure-5.7.1 LLEBF.png


text: File:FE1 Team6 Figure- 5.7.2 Coordinate change.png


text: File:FE1 Team6 Figure 5.7.3 4node.png

title: File:FE1 Team6 Figure 5.7.3 4node.png


text: File:FE1 Team6 Figure 5.7.4 6nodes.png

title: File:FE1 Team6 Figure 5.7.4 6nodes.png


text: File:FE1 Team6 Figure -5.7.5 4nodes.png

title: File:FE1 Team6 Figure -5.7.5 4nodes.png


text: File:FE1 Team6 Figure - 5.7.6 6nodes.png


text:
User:Eml5526.s11.team2.stewart/HW3

clear all; clc; format long syms x Uh; U = -118/27*log((2+3*x)/5) + (20*x-15*x^2+139)/36; % Exact Solution m = 2;                                            % Global Equidistant Nodes d_0 = 4; x_m = zeros(m); for i=1:m x_m(i)=(i-1)/(m-1); end

% Basis Function Vector b_i = sym(ones(m,1)); for i=1:m for j=1:m b_i(i)=(x-1)^(i); end end

% Construct K and F Kff=ones(m,m); Ff=ones(m,1); for i=1:m for j=1:m Kff(i,j)=int(diff(b_i(i))*(2+3*x)*diff(b_i(j)),0,1); end Ff(i)=int(b_i(i)*5*x,0,1) + subs((12+18*x)*b_i(i),x,0); end

d = zeros(m,1); Kff            % Lazy way of displaying Kff Ff             % Lazy way of displaying Ff d = Kff\Ff;     % Solve for d i=1,2,...,m d = [d_0;d]    % Prepend d_0 b_i = [1;b_i]  % Prepend b_0

Uh = d'*b_i    % This come out simplified with fractions. % You can just look at b_i and d to write out Uh

title: User:Eml5526.s11.team2.stewart/HW3


title: File:Eml5526.s11.team2.matlab.png


text:
User:Egm6341.s11.team2.zou/HW5

=Problem 5.3 Cauchy Distribution and Normal Distribution=

From the lecture slide Mtg26

Given:
The Cauchy distribution:

The Normal or Gauss distribution:

and the definition of quartiles $$\displaystyle {{Q}_{1}}$$, $$\displaystyle {{Q}_{2}}$$ and $$\displaystyle {{Q}_{3}}$$:

Find:
a) Three quartiles $$\displaystyle \left\{ Q_{1}^{C},Q_{3}^{C} \right\}$$ for Cauchy distribution

b) Three quartiles $$\displaystyle \left\{ Q_{1}^{G},Q_{3}^{G} \right\}$$ for Normal distribution

c) Let $$\displaystyle {{x}_{0}}=\mu =0$$ and $$\displaystyle {{\gamma }^{C}}=1$$ where $$\displaystyle {{\gamma }^{C}}$$ is the half width of $$\displaystyle C({{x}_{0}},\gamma )$$. Find $$\displaystyle {{\sigma }^{1}}$$ such that the half width of normal distribution $$\displaystyle {{\gamma }^{G}}=1$$. Then plot $$\displaystyle C(0,1)$$ and $$\displaystyle N(0,{{\sigma }^{1}})$$

d) Find $$\displaystyle \left\{ Q_{1}^{C},Q_{3}^{C} \right\}$$ for $$\displaystyle C(0,1)$$ and $$\displaystyle \left\{ Q_{1}^{G},Q_{3}^{G} \right\}$$ for $$\displaystyle N(0,{{\sigma }^{1}})$$. Then plot them with comments on results.

a) Three quartiles for Cauchy distribution
From the WolframAlpha we have got the general expression for the quartiles $$\displaystyle Q_{1}^{C}$$ and $$\displaystyle Q_{3}^{C}$$:

We can prove this by finding the cumulative distribution function of Cauchy distribution,

Then,

Similarly,

b) Three quartiles for Normal distribution
From the WolframAlpha we can find the quartiles for the normal distribution,

And also we can find the cumulative of normal distribution,

Then we can check the validity of the quartiles given.

c) Compare Cauchy and Gauss distribution with same half width.
With the information given we have our Cauchy distribution with half width equals 1,

To find a normal distribution we need to know the maximum value of

Hence,

Then we must find the x coordinates in the half width nodes.

The difference between the 2 roots should be twice the half width which is 2, hence,

which means the normal distribution is,

The plots of them can be achieved by WolframAlpha

d) Find quartiles for particular Cauchy and Normal distribution. Then plot them with comments on results.
By the formula from part a) and part b) we have,

The following figure is plotted by Adam Franklin,

Corrsponding Matlab Code

Comments

From the results above we can see that the range from $$\displaystyle Q_{1}^{C}$$ to $$\displaystyle Q_{3}^{C}$$ is significantly larger than that from $$\displaystyle Q_{1}^{G}$$ to $$\displaystyle Q_{3}^{G}$$. The probability outside (-0.2865, 0.2865) decrease drastically to very small in normal distribution while in Cauchy distribution this is until (-1, 1). This means the Cauchy distribution allow more probability on the large outcome to occur.

title: User:Egm6341.s11.team2.zou/HW5


text: File:HW5.1 poly.png

title: File:HW5.1 poly.png


text: File:FE1 Team6 Fig- 5.2.1.jpg

title: File:FE1 Team6 Fig- 5.2.1.jpg


text: File:HW5.1.poly error.png

title: File:HW5.1.poly error.png


text:
Honours thesis in psychology/Proposal

Overview
Honours theses often require a research proposal which is formally reviewed because:
 * 1) Developing a research proposal and getting peer review is a critical step in the scientific method.
 * 2) It is better to find out about potential research design flaws early on (e.g., from proposal reviewers), while they can still be addressed, than to find out later (e.g., from examiners).

Format
A research proposal may take different forms e.g.:
 * 1) presentation at a meeting of staff and/or fellow students
 * 2) poster presentation

Expect to speak for 5 to 20 minutes and to accompany your talk with some audiovisual material (such as slides). Also expected to answer questions about your proposal.

For example, for psychology Honours theses at the University of Canberra's poster, present:
 * 1) Electronically by creating a new discussion thread post with an attached file via the LMS.
 * 2) Poster or Slides which provide easy-to-read, key details about about the study to help reviewers understand the purpose and design and to inform critique.
 * 3) A 5-minute verbal précis of the proposal to reviewers and interested fellow students.
 * 4) Copies of relevant additional information (e.g., draft survey).

Reviewers will discuss your proposal with you and provide some written comments for further discussion with your supervisor. Obtain at least two reviews from academic staff. Fellow students may provide additional feedback. If appropriate, make adjustments to the study's focus and design.

Sections
These sections are recommended for research proposals (check with your supervisor for further advice):
 * 1) Title page
 * 2) Working title
 * 3) Student name
 * 4) Supervisor name
 * 5) Date
 * 6) Introduction: A brief review of the literature and justification for why the topic is important.
 * 7) Aims & Hypotheses: Concise statement of research question(s) (aims) and briefly how these aims fit into the literature area (justification of your aims). Clearly state the study’s hypotheses (Note: Hypotheses need to be testable – see also Analyses).
 * 8) Method:
 * 9) Design: Describe the research design (e.g., experimental, quasi-experimental, or non-experimental, between-subjects or within-subjects, etc.). Consider potential confounds and how they will be dealt with (internal validity).
 * 10) Participants: Consider the target population, sampling frame, sampling technique, anticipated return rate, sample size and power (demonstrate how the sample size is adequate for your design). Note whether you or your supervisor have a dependent relationship with this target population. Describe how participants will be recruited.
 * 11) Materials/Measures: Describe how the independent and dependent measures are to be operationalised and the psychometric properties of proposed measures.
 * 12) Procedure: How will the study be carried out? How will the data be collected? What steps will be followed with participants? What will be done with the data?
 * 13) Ethical issues: Summarise potential ethical issues and explain how they will be dealt with. Consider whether approval is required from the University’s Committee for Ethics in Human Research or any other organisation.
 * 14) Costs: Provide details of any project costs, such as equipment, and how these will be paid. Where possible, students are encouraged to use free materials.
 * 15) Analyses: Summarise the planned data analytical techniques for addressing the hypotheses.
 * 16) Timeline: Outline key project milestones and their due dates, including ethical clearances. See timeline for more information.
 * 17) References
 * 18) Appendices: e.g., a copy of the instrumentation to be used, informed consent statement

title: Honours thesis in psychology/Proposal


title: File:HW.5.2.poly.png


text: File:HW5.2.poly error.png

title: File:HW5.2.poly error.png


title: File:Fea1.jpg


title: File:Fea2.jpg


text: File:FE1 Team6 Fig- 5.2.2.jpg

title: File:FE1 Team6 Fig- 5.2.2.jpg


text: File:FE1 Team6 Fig- 5.2.2 .jpg

title: File:FE1 Team6 Fig- 5.2.2 .jpg


text:
User:Egm6321.f10.team2.oztekin/Newwwww

=Problem 5.8:Using LLEBF to solve G1DM1.0/D1=

Given
G1DM1.0/D1 : Model for elastic bar problem was given in meeting 9.

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

$$  \displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\left [ (2+3x)\frac{\mathrm{d} u}{\mathrm{d} x} \right ]+5x=0 $$     (8.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle u(1)=4 $$     (8.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle \frac{\mathrm{d} u(x=0)}{\mathrm{d} x}=-3 $$     (8.3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle \int_{\Omega }\frac{\mathrm{d} w}{\mathrm{d} x}AE\frac{\mathrm{d} u}{\mathrm{d} x}dx-\left ( wAt \right )_{\Gamma _{h}}-\int_{\Omega }wbdx=0 \Rightarrow \Omega =]0,1[ $$     (8.4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

We have to manipulate weak form in order to use for element matrices. As to element mesh we use weak form for each elements and then we can take summation for all elements to get global matrices.

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

$$  \displaystyle \sum\limits_{e=1}^{nel}{{{w}^{eT}}\left\{ \int\limits_{{{x}_{1}}^{e}}^{{{x}_{2}}^{e}}{{{B}^{eT}}{{A}^{e}}{{E}^{e}}{{B}^{e}}dx{{d}^{e}}-\int\limits_{{{x}_{1}}^{e}}^{x_{2}^{e}}{{{N}^{eT}}bdx-{{({{N}^{eT}}{{A}^{e}}\overline{t})}_{x=0}}}} \right\}=0} $$     (8.5)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Approximate functions gives as in F&B Book,

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

$$  \displaystyle u^{h}=\sum_{e=1}^{nel}N^{e}d^{e}=\left ( \sum_{e=1}^{nel}N^{e}L^{e}\right )\mathbf{d} $$     (8.6)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle w^{h}=\sum_{e=1}^{nel}N^{e}w^{e}=\left ( \sum_{e=1}^{nel}N^{e}L^{e}\right )\mathbf{w} $$     (8.7)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Where N is called as 'shape functions' and L is called 'gather matrices'.

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

$$  \displaystyle N^{e}=\frac{1}{l^{e}}[\begin{matrix} x_{2}^{e}-x & x-x_{1}^{e} \end{matrix}] $$     (8.8)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle B^{e}=\frac{1}{l^{e}}[\begin{matrix} -1 & 1 \end{matrix}] $$     (8.9)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Analytical Solution

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

$$  \displaystyle \int \left ( (2+3x)\frac{\mathrm{d} u}{\mathrm{d} x} \right )dx=-\int (5x)dx+c_{1} $$     (8.10)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle 2\frac{\mathrm{d} u}{\mathrm{d} x}(x=0)=0+c_{1} \Rightarrow c_{1}=-6 $$     (8.11)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle u=-\frac{5}{2}\int \frac{x^2}{2+3x}dx-\int \frac{6}{2+3x}dx+c_{2} $$     (8.12)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle -\frac{5}{2}\left [ \frac{1}{54}(9x^2-12x+8log(3x+2)-12) \right ]-\left [ \frac{1}{3}log(3x+2) \right ]+c_{2} $$     (8.13)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Using other boundry condition ;

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

$$  \displaystyle c_{2}=\frac{119}{36}+\frac{64}{27}log(5) $$     (8.14)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle u(x)=-\frac{5}{12}x^2+\frac{5}{9}x-\frac{64}{27}log(2+3x)+\frac{139}{36}+\frac{64}{27}log(5) $$     (8.15)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

FEA solutions
When evaluating unknowns on distributed nodes our shape functions must meet some conditions.Our approximate functions must be complete in each element. As saying completeness we are saying that by using our nodal values we must be able to identify coefficients in our approximate solutions. After some manipulation we conclude with shape functions. For each nodes we have shape functions and for middle nodes shape functions come from adjacent elements.

For example two element mesh ;
 * {| style="width:100%" border="0"

$$  \displaystyle N=N^{(1)}L^{(1)}+N^{(2)}L^{(2)}=[\begin{matrix} N_{1}^{(1)} & N_{2}^{(1)}+N_{1}^{(2)} & N_{2}^{(2)} \end{matrix}]
 * style="width:95%" |
 * style="width:95%" |

$$     (8.16)
 *
 * }

For this example at x1 (first node) only N1 will be 1. Others will be zero. So we verify CBS.

Using 4 elements

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

$$  \displaystyle \Omega =]0,1[ $$
 * style="width:95%" |
 * style="width:95%" |

 * }

Lets find shape functions for 4 elements.

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

$$  \displaystyle\begin{align} &N^{(1)}=\frac{1}{0.25}[\begin{matrix} 0.25-x & x \end{matrix}]\\ &N^{(2)}=\frac{1}{0.25}[\begin{matrix} 0.5-x & x-0.25 \end{matrix}]\\ &N^{(3)}=\frac{1}{0.25}[\begin{matrix} 0.75-x & x-0.5 \end{matrix}]\\ &N^{(4)}=\frac{1}{0.25}[\begin{matrix} 1-x & x-0.75 \end{matrix}] \end{align} $$     (8.17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

These shape functions are useful for us to make connection between nodes. After defining these tools now we can find stiffness matrices for all elements. Then after finding all of stiffness matrices for elements we can construct global stiffness matrix.

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

$$  \displaystyle K^{e}=\int_{x_{1}^{e}}^{x_{2}^{e}}B^{eT}A^{e}E^{e}B^{e}dx $$     (8.18)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle\begin{align} {{K}^{e}}&=\int\limits_{x_{1}^{e}}^{x_{2}^{e}}{\frac{1}\left[ \begin{matrix} -1 \\   1  \\ \end{matrix} \right]}(2+3x)\frac{1}[\begin{matrix} -1 & 1 \\ \end{matrix}]dx \\ &=\frac{1}\left[ \begin{matrix} 1 & -1 \\   -1 & 1  \\ \end{matrix} \right]\left[ 2(x_{2}^{e}-x_{1}^{e})+\frac{3}{2}(x_{2}^{e2}-x_{1}^{e2}) \right] \end{align} $$     (8.19)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle {{K}^{e}}=\frac{1}\left[ \begin{matrix} 1 & -1 \\   -1 & 1  \\ \end{matrix} \right]\left[ 2+\frac{3}{2}(x_{2}^{e}+x_{1}^{e}) \right] $$     (8.20)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle K^{(1)}=\begin{bmatrix} 9.5 & -9.5\\ -9.5& 9.5 \end{bmatrix} $$     (8.21)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle K^{(2)}=\begin{bmatrix} 12.5 & -12.5\\ -12.5& 12.5 \end{bmatrix} $$     (8.22)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle K^{(3)}=\begin{bmatrix} 15.5 & -15.5\\ -15.5& 15.5 \end{bmatrix} $$     (8.23)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle K^{(4)}=\begin{bmatrix} 18.5 & -18.5\\ -18.5& 18.5 \end{bmatrix} $$     (8.24)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We can construct global stiffness matrix now. This will be distribution of all four element stiffness matrices into appropriate places. For our convenience i will use direct method. But it is useful using below equation for computing purposes.

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

$$  \displaystyle K=\sum_{e=1}^{nel}L^{eT}K^{e}L^{e}=L^{1T}K^{1}L^{1}+L^{2T}K^{2}L^{2}+L^{3T}K^{3}L^{3}+L^{4T}K^{4}L^{4} $$     (8.25)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle K=\left[ \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} 9.5 \\   -9.5  \\ \end{matrix}  \\ 0 \\   0  \\   0  \\ \end{matrix} & \begin{matrix} \begin{matrix} -9.5 \\   22  \\ \end{matrix}  \\ -12.5 \\   0  \\   0  \\ \end{matrix}  \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 \\   -12.5  \\ \end{matrix}  \\ 28 \\   \begin{matrix} -15.5 \\   0  \\ \end{matrix}  \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 \\   0  \\ \end{matrix}  \\ -15.5 \\   34  \\   -18.5  \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 \\   0  \\ \end{matrix}  \\ 0 \\   -18.5  \\   18.5  \\ \end{matrix}  \\ \end{matrix} \right]
 * style="width:95%" |
 * style="width:95%" |

$$     (8.26)
 *

 * }

The next step we have to find force matrices. Which ha two components. Lets first find force matrix at natural boundary condition and second distributed force matrix on our domain.

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

$$  \displaystyle f_{\Gamma _{h}}=-(N^{eT}A^{e}b)_{\Gamma _{h}^{e}}=-N^{eT}(x_{1})(2+3x)_{x=0}(-3)=6N^{eT}(x_{1}) $$     (8.27)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle f_{{\Gamma }^{(1)}}=6\left[ \begin{matrix} N_{1}^{(1)}({{x}_{1}}) \\ N_{2}^{(1)}({{x}_{2}}) \\ \end{matrix} \right]=\left[ \begin{matrix} 6 \\   0  \\ \end{matrix} \right]
 * style="width:95%" |
 * style="width:95%" |

$$     (8.28)
 *

 * }

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

$$  \displaystyle\begin{align} &f_{{\Gamma }^{(2)}}=f_{{\Gamma }^{(3)}}=f_{{\Gamma }^{(4)}}=6\left[ \begin{matrix} N_{1}^{(e)}({{x}_{1}}) \\ N_{2}^{(e)}({{x}_{2}}) \\ \end{matrix} \right]=\left[ \begin{matrix} 0 \\   0  \\ \end{matrix} \right] &for & e=2,3,4 \end{align} $$     (8.29)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle\begin{align} f_{\Gamma _{h}} &=\sum_{e=1}^{nel}L^{eT}f_{\Gamma _{h}}^{e} \\ &=\left[ \begin{matrix} \begin{matrix} 6 \\   0  \\ \end{matrix}  \\ 0 \\   0  \\   0  \\ \end{matrix} \right]
 * style="width:95%" |
 * style="width:95%" |

\end{align} $$     (N)
 *

 * }

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

$$  \displaystyle\begin{align} f_{\Omega }^{e} & =\int_{x_{1}^{e}}^{x_{2}^{e}}N^{eT}b(x)dx\\ &=\int\limits_{x_{1}^{e}}^{x_{2}^{e}}{\frac{1}\left[ \begin{matrix} x_{2}^{e}-x \\ x-x_{1}^{e} \\ \end{matrix} \right]}5xdx \\ &=\frac{5}\left[ \begin{matrix} \frac{x_{2}^{e}(x_{2}^{e2}-x_{1}^{e2})}{2}-\frac{(x_{2}^{e3}-x_{1}^{e3})}{3} \\ \frac{(x_{2}^{e3}-x_{1}^{e3})}{3}-\frac{x_{1}^{e}(x_{2}^{e2}-x_{1}^{e2})}{2} \\ \end{matrix} \right]
 * style="width:95%" |
 * style="width:95%" |

\end{align} $$     (8.30)
 *

 * }

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

$$  \displaystyle f_{\Omega }^{(1)}=\left[ \begin{matrix} 0.05208334 \\   0.10416666  \\ \end{matrix} \right] $$     (8.31)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle f_{\Omega }^{(2)}=\left[ \begin{matrix} 0.2083334 \\   0.2604166  \\ \end{matrix} \right] $$     (8.32)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle f_{\Omega }^{(3)}=\left[ \begin{matrix} 0.36458334 \\   0.41666666  \\ \end{matrix} \right] $$     (8.33)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle f_{\Omega }^{(4)}=\left[ \begin{matrix} 0.52083334 \\   0.57291666  \\ \end{matrix} \right] $$     (8.34)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle\begin{align} f_{\Omega }&=\sum_{e=1}^{4}L^{eT}f_{\Omega }^{e} &\left[ \begin{matrix} \begin{matrix} 0.05208334 \\   0.3125  \\ \end{matrix}  \\ 0.625 \\   0.9375  \\   0.57291666  \\ \end{matrix} \right]
 * style="width:95%" |
 * style="width:95%" |

\end{align} $$     (8.35)
 *

 * }

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

$$  \displaystyle \left[ \begin{matrix} \begin{matrix} 9.5 \\   -9.5  \\ \end{matrix} & \begin{matrix} -9.5 \\   22  \\ \end{matrix} & \begin{matrix} 0 \\   -12.5  \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 & 0 \\ \end{matrix}  \\ \begin{matrix} 0 & 0 \\ \end{matrix}  \\ \end{matrix} \\ 0 & -12.5 & 28 & \begin{matrix} -15.5 & 0 \\ \end{matrix}  \\ 0 & 0 & -15.5 & \begin{matrix} 34 & -18.5 \\ \end{matrix}  \\ 0 & 0 & 0 & \begin{matrix} -18.5 & 18.5 \\ \end{matrix}  \\ \end{matrix} \right]\left[ \begin{matrix} \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ \end{matrix} \\ {{d}_{3}} \\ {{d}_{4}} \\ 4 \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} 0.05208334 \\   0.3125  \\ \end{matrix}  \\ 0.625 \\   0.9375  \\   0.57291666  \\ \end{matrix} \right]+\left[ \begin{matrix} \begin{matrix} 6 \\   0  \\ \end{matrix}  \\ 0 \\   0  \\   0  \\ \end{matrix} \right]+\left[ \begin{matrix} \begin{matrix} 0 \\   0  \\ \end{matrix}  \\ 0 \\   0  \\   {{r}_{1}}  \\ \end{matrix} \right]
 * style="width:95%" |
 * style="width:95%" |

$$     (8.36)
 *

 * }

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

$$  \displaystyle \left[ \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ {{d}_{3}} \\ {{d}_{4}} \\ \end{matrix} \right]=\left[ \begin{matrix} 2.0257 \\   1.3886  \\   0.8794  \\   0.4285  \\ \end{matrix} \right]
 * style="width:95%" |
 * style="width:95%" |

$$     (8.37)
 *

 * }

Comparing results with exact ones at the nodes;

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

$$  \displaystyle\begin{align} &\left[ \begin{matrix} 2.0257 \\   1.3886  \\   0.8794  \\   0.4285  \\ \end{matrix} \right]+\left[ \begin{matrix} 4 \\   4  \\   4  \\   4  \\ \end{matrix} \right]=\left[ \begin{matrix} 6.0257 \\   5.3886  \\   4.8794  \\   4.4285  \\ \end{matrix} \right]
 * style="width:95%" |
 * style="width:95%" |

&{{d}_{exact}}=\left[ \begin{matrix} 6.0331 \\   5.3911  \\   4.8802  \\   4.4286  \\ \end{matrix} \right]

\end{align} $$     (8.38)
 *

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

$$  \displaystyle 4 elements $$
 * style="width:95%" |
 * style="width:95%" |

 * }

Using 6 elements
Shape functions for 6 elements
 * {| style="width:100%" border="0"

$$  \displaystyle\begin{align} & N^{(1)}=\frac{1}{0.1666}[\begin{matrix} 0.1666-x & x \end{matrix}]\\ &N^{(2)}=\frac{1}{0.1666}[\begin{matrix} 0.3333-x & x-0.16666 \end{matrix}]\\ & N^{(3)}=\frac{1}{0.1666}[\begin{matrix} 0.5-x & x-0.3333 \end{matrix}]\\ &N^{(4)}=\frac{1}{0.1666}[\begin{matrix} 0.6666-x & x-0.5 \end{matrix}]\\ & N^{(5)}=\frac{1}{0.1666}[\begin{matrix} 0.8332-x & x-0.6666 \end{matrix}]\\ & N^{(6)}=\frac{1}{0.1666}[\begin{matrix} 1-x & x-0.8332 \end{matrix}]\\
 * style="width:95%" |
 * style="width:95%" |

\end{align} $$     (8.39)
 *

 * }

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

$$  \displaystyle \left[ \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} 13.5048 & -13.5048 & 0 & 0 \\ \end{matrix} & 0 & 0 & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} -13.5048 & 30.0105 & -16.5057 & 0 \\ \end{matrix} & 0 & 0 & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & -16.5057 & 36.0132 & -19.5075 \\ \end{matrix} & 0 & 0 & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 & -19.5075 & 42.0157 \\ \end{matrix} & -22.5084 & 0 & 0  \\ \end{matrix} \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 & 0 & -22.5084 \\ \end{matrix} & 48.0164 & -25.508 & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 & 0 & 0 \\ \end{matrix} & -25.508 & 54.008 & -28.5  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 & 0 & 0 \\ \end{matrix} & 0 & -28.5 & 28.5  \\ \end{matrix} \\ \end{matrix} \right]\left[ \begin{matrix} \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ {{d}_{3}} \\ {{d}_{4}} \\ \end{matrix} \\ {{d}_{5}} \\ {{d}_{6}} \\ 4 \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} 6.0232199942 \\   0.13890282  \\   0.277972087  \\   0.416641681  \\ \end{matrix}  \\ 0.555277743 \\   0.694970317  \\   0.394287575  \\ \end{matrix} \right]+\left[ \begin{matrix} \begin{matrix} 0 \\   0  \\   0  \\   0  \\ \end{matrix}  \\ 0 \\   0  \\   {{r}_{1}}  \\ \end{matrix} \right]
 * style="width:95%" |
 * style="width:95%" |

$$     (8.40)
 *

 * }

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

$$  \displaystyle \left[ \begin{matrix} \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ {{d}_{3}} \\ \end{matrix} \\ {{d}_{4}} \\ {{d}_{5}} \\ {{d}_{6}} \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} 2.0292 \\   1.5831  \\   1.2098  \\ \end{matrix}  \\ 0.8797 \\   0.575  \\   0.2845  \\ \end{matrix} \right]
 * style="width:95%" |
 * style="width:95%" |

$$     (8.41)
 *

 * }

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

$$  \displaystyle\begin{align} & \left[ \begin{matrix} \begin{matrix} 2.0292 \\   1.5831  \\   1.2098  \\ \end{matrix}  \\ 0.8797 \\   0.575  \\   0.2845  \\ \end{matrix} \right]+\left[ \begin{matrix} \begin{matrix} 4 \\   4  \\   4  \\ \end{matrix}  \\ 4 \\   4  \\   4  \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} 6.0292 \\   5.5831  \\   5.2098  \\ \end{matrix}  \\ 4.8797 \\   4.575  \\   4.2845  \\ \end{matrix} \right]
 * style="width:95%" |
 * style="width:95%" |

& {{d}_{exact}}=\left[ \begin{matrix} \begin{matrix} 6.0331 \\   5.5853  \\   5.2109  \\ \end{matrix}  \\ 4.8802 \\   4.5753  \\   4.2847  \\ \end{matrix} \right]

\end{align} $$     (8.42)
 *

 * }

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

$$  \displaystyle 6 elements $$
 * style="width:95%" |
 * style="width:95%" |

 * }

Using 8 elements
Shape functions for 8 elements
 * {| style="width:100%" border="0"

$$  \displaystyle\begin{align} & N^{(1)}=\frac{1}{0.125}[\begin{matrix} 0.125-x & x \end{matrix}]\\ &N^{(2)}=\frac{1}{0.125}[\begin{matrix} 0.25-x & x-0.125 \end{matrix}]\\ & N^{(3)}=\frac{1}{0.125}[\begin{matrix} 0.375-x & x-0.25 \end{matrix}]\\ &N^{(4)}=\frac{1}{0.125}[\begin{matrix} 0.5-x & x-0.375 \end{matrix}]\\ & N^{(5)}=\frac{1}{0.125}[\begin{matrix} 0.625-x & x-0.5 \end{matrix}]\\ & N^{(6)}=\frac{1}{0.125}[\begin{matrix} 0.75-x & x-0.625 \end{matrix}]\\ & N^{(7)}=\frac{1}{0.125}[\begin{matrix} 0.875-x & x-0.75 \end{matrix}]\\ & N^{(8)}=\frac{1}{0.125}[\begin{matrix} 1-x & x-0.875 \end{matrix}]\\ \end{align} $$     (8.43)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \left[ \begin{matrix} 17.5 & -17.5 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\   -17.5 & 38 & -20.5 & 0 & 0 & 0 & 0 & 0 & 0  \\   0 & -20.5 & 44 & -23.5 & 0 & 0 & 0 & 0 & 0  \\   0 & 0 & -23.5 & 50 & -26.5 & 0 & 0 & 0 & 0  \\   0 & 0 & 0 & -26.5 & 56 & -29.5 & 0 & 0 & 0  \\   0 & 0 & 0 & 0 & -29.5 & 62 & -32.5 & 0 & 0  \\   0 & 0 & 0 & 0 & 0 & -32.5 & 68 & -35.5 & 0  \\   0 & 0 & 0 & 0 & 0 & 0 & -35.5 & 74 & -38.5  \\   0 & 0 & 0 & 0 & 0 & 0 & 0 & -38.5 & 38.5  \\ \end{matrix} \right]\left[ \begin{matrix} \begin{matrix} \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ {{d}_{3}} \\ \end{matrix} \\ {{d}_{4}} \\ {{d}_{5}} \\ {{d}_{6}} \\ \end{matrix} \\ {{d}_{7}} \\ {{d}_{8}} \\ {{d}_{9}} \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} 6.01302084 \\   0.078125  \\ \end{matrix}  \\ 0.15625 \\ \end{matrix}  \\ 0.234375 \\   0.31250004  \\   0.390625  \\ \end{matrix}  \\ 0.46874373 \\   0.546875  \\   0.29947916  \\ \end{matrix} \right]+\left[ \begin{matrix} \begin{matrix} \begin{matrix} 0 \\   0  \\   0  \\ \end{matrix}  \\ 0 \\   0  \\   0  \\ \end{matrix}  \\ 0 \\   0  \\   {{r}_{1}}  \\ \end{matrix} \right]
 * style="width:95%" |
 * style="width:95%" |

$$     (8.44)
 *

 * }

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

$$  \displaystyle \left[ \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ \end{matrix} \\ {{d}_{3}} \\ \end{matrix} \\ {{d}_{4}} \\ {{d}_{5}} \\ \end{matrix} \\ {{d}_{6}} \\ {{d}_{7}} \\ {{d}_{8}} \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} \begin{matrix} 2.0312 \\   1.6876  \\ \end{matrix}  \\ 1.3904 \\   1.1246  \\   0.8800  \\ \end{matrix}  \\ 0.6497 \\   0.4286  \\   0.213  \\ \end{matrix} \right]
 * style="width:95%" |
 * style="width:95%" |

$$     (8.45)
 *

 * }

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

$$  \displaystyle\begin{align} &\left[ \begin{matrix} \begin{matrix} \begin{matrix} 2.0312 \\   1.6876  \\ \end{matrix}  \\ 1.3904 \\   1.1246  \\   0.8800  \\ \end{matrix}  \\ 0.6497 \\   0.4286  \\   0.213  \\ \end{matrix} \right]+\left[ \begin{matrix} \begin{matrix} \begin{matrix} 4 \\   4  \\ \end{matrix}  \\ 4 \\   4  \\   4  \\ \end{matrix}  \\ 4 \\   4  \\   4  \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} \begin{matrix} 6.0312 \\   5.6876  \\ \end{matrix}  \\ 5.3904 \\   5.1246  \\   4.88  \\ \end{matrix}  \\ 4.6497 \\   4.4286  \\   4.213  \\ \end{matrix} \right]
 * style="width:95%" |
 * style="width:95%" |

&{{d}_{exact}}=\left[ \begin{matrix} \begin{matrix} \begin{matrix} 6.0331 \\   5.6866  \\ \end{matrix}  \\ 5.3911 \\   5.1249  \\   4.8802  \\ \end{matrix}  \\ 4.6498 \\   4.4286  \\   4.213  \\ \end{matrix} \right]

\end{align} $$     (8.46)
 *

 * }

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

$$  \displaystyle 8 elements $$
 * style="width:95%" |
 * style="width:95%" |

 * }

title: User:Egm6321.f10.team2.oztekin/Newwwww


title: File:Eml5526.s11.team2.equ.png


title: File:Fe1.s11.team6.kurth.hw5.fig1.jpg


title: File:Eml5526.s11.team2.equ2.png


text:
User:Michaeljabal

In this page I'll summarize my understanding of various theories behind eye tracking technologies.

title: User:Michaeljabal


title: File:5-3a.gif


text: File:Eml5526.s11.team4.simko.hw5.problem3.fig3.png


text: File:Eml5526.s11.team7.hw5.7 1.jpg

title: File:Eml5526.s11.team7.hw5.7 1.jpg


text: File:Eml5526.s11.team7.hw5.7 2.jpg

title: File:Eml5526.s11.team7.hw5.7 2.jpg


title: File:Eml5526.s11.team6.hw5.5.1.jpg


title: File:Eml5526.s11.team6.hw5.5.2.jpg


title: File:Eml5526.s11.team6.hw5.5.3.jpg


text: File:FEA HW 5 1 E 2.jpg

title: File:FEA HW 5 1 E 2.jpg


text: File:FEA HW 5 1 E 3.jpg

title: File:FEA HW 5 1 E 3.jpg


title: File:Eml5526.s11.team6.hw5.5.4.jpg


text: File:FEA HW 5 1 F 2.jpg

title: File:FEA HW 5 1 F 2.jpg


title: File:Eml5526.s11.team6.hw5.5.5.jpg


text: File:FEA HW 5 1 F 3.jpg

title: File:FEA HW 5 1 F 3.jpg


title: File:Eml5526.s11.team6.hw5.5.6.jpg


text: File:FEA HW 5 1 P 2.jpg

title: File:FEA HW 5 1 P 2.jpg


text: File:FEA HW 5 1 P 3.jpg

title: File:FEA HW 5 1 P 3.jpg


text: File:FEA HW 5 2 E 2.jpg

title: File:FEA HW 5 2 E 2.jpg


text: File:FEA HW 5 2 E 3.jpg

title: File:FEA HW 5 2 E 3.jpg


text: File:FEA HW 5 2 F 2.jpg

title: File:FEA HW 5 2 F 2.jpg


text: File:FEA HW 5 2 F 3.jpg

title: File:FEA HW 5 2 F 3.jpg


text: File:FEA HW 5 2 P 2.jpg

title: File:FEA HW 5 2 P 2.jpg


text: File:FEA HW 5 2 P 3.jpg

title: File:FEA HW 5 2 P 3.jpg


title: File:LLEBF(1).png


text: File:Eml5526.s11.team4.simko.hw5.problem3.fig4.png


title: File:LLEBF(2).png


title: File:Eml5526.s11.team2.polyplot.png


title: File:Eml5526.s11.team2.polyerror.png


title: File:Hw5.6(1).png


title: File:Hw5.6(2).png


title: File:Hw5.6(3).png


text: File:Egm6341.s11.team5.Hw5 3.jpg

title: File:Egm6341.s11.team5.Hw5 3.jpg


title: File:Hw5.6(4).png


text: File:Egm6341.s11.team5.Hw5 4.jpg

title: File:Egm6341.s11.team5.Hw5 4.jpg


title: File:5-6a.png


title: File:5-6b.png


title: File:5-6c.png


title: File:5-6d.png


title: File:Fe1.s11.tm7.HW5.exp.jpg


title: File:Fe1.s11.tm7.HW5.experror.jpg


text:
User:Eml5526.s11.team4.yang/hw5

=HW 5.3 problem: Solving G1DM1.0/D1b using LIBF=

Refer to lecture slide [[media:fe1.s11.mtg29.djvu|mtg-29]] for the problem statement.

Given
G1DM1.0/D1 (Image from Charles Cook, HW1.1) $$\Omega=\left\{0,1\right\}$$ $$\Gamma_{g}=\left\{0\right\}, g = 4$$ $$\Gamma _{h}=\left\{1\right\}, h = 12$$ $$a_{2}\left\{x\right\}=2+3x$$ $$f\left\{x,t\right\}=5x$$ $$\frac{\partial^{2} u}{\partial t^{2}}=0$$

Objective
1. Explain how LIBF are used as CBS 2. Plot all LIBF used 3. Use matlab quad, WA, to int. 4. plot $$u_{m}^{h}$$ VS u, $$u_{m}^{h}(0.5)-u(0.5)$$ VS m

solution
1. According the definition of constraint breaking solution(CBS), we can prove the LIBF can satisfy the CBS.

 * $$l_{m}^{i}(x)=\prod\limits_{j=1,j\ne i}^{m}{\frac{(x-{{x}_{j}})}{({{x}_{i}}-{{x}_{j}})}}$$

where,
 * $${{x}_{1}}=0,{{x}_{2}}=1/3,{{x}_{3}}=2/3,{{x}_{4}}=1$$.

At the boundary of $$\beta $$ , it can be solved that: $$\begin{align} & \beta =0, \\ & l_{m}^{1}(x=0)=1; \\ & l_{m}^{2}(x=0)=l_{m}^{3}(x=0)=l_{m}^{4}(x=0)=0 \\ \end{align}$$ So, these four basis function can satisfy the CBS. The procedure is the same for m=6, m=8 here. 2. Plot all the basis used for m=4, m=6, m=8, respectively. m=4 m=6 m=8 3. The exact solution is obtained by following steps. $$\frac{\partial}{\partial x}\left[(2+3x)\frac{\partial u}{\partial x}\right]+5x=0$$ Integrate the above equation, we obtain that: $$\left[(2+3x)\frac{\partial u}{\partial x}\right]+\frac{5x^{2}}{2}+C_1=0$$ Applying the natural boundary condition: $$\Gamma _{h}=\left\{0\right\}, h = 12$$ which means: $${{a}_{2}}(x)\centerdot \frac{\partial u}{\partial x}(x=1)=(2+3x)\centerdot \frac{\partial u}{\partial x}(x=1)=12$$ Then, we obtain that: $$\mathbf{C_1 = -29/2}$$ Then, we get: $$\left[(2+3x)\frac{\partial u}{\partial x}\right]+\frac{5x^{2}}{2}-29/2=0$$ Using WA to solve the equation, and applying the essential boundary condition yields the exact solution: $$u=-5{{x}^{2}}/12+5x/9+241\centerdot log(3x+2)/54-241\centerdot log(2)/54+4;$$ 4. m=4, the exact solution and approximate solution using LIBF are compared in the following figure: The according matlab code is: m=6, the exact solution and approximate solution using LIBF are compared in the following figure: The according matlab code is:

m=8, the exact solution and approximate solution using LIBF are compared in the following figure: The according matlab code is: The error is decreased by increasing the nodes of basis function(LIBF), which can be clearly in the following figure: The according matlab code is:

This problem was solved by shengfeng yang

=HW 5.4 problem: Solving G1DM1.0/D1 using LIBF=

Refer to lecture slide [[media:fe1.s11.mtg29.djvu|mtg-29]] for the problem statement.

Given
G1DM1.0/D1 (Image from Charles Cook, HW1.1) $$\Omega=\left\{0,1\right\}$$ $$\Gamma_{g}=\left\{1\right\}, g = 4$$ $$\Gamma _{h}=\left\{0\right\}, h = 12$$ $$a_{2}\left\{x\right\}=2+3x$$ $$f\left\{x,t\right\}=5x$$ $$\frac{\partial^{2} u}{\partial t^{2}}=0$$

Objective
1. Explain how LIBF are used as CBS 2. Plot all LIBF used 3. Use matlab quad, WA, to int. 4. plot $$u_{m}^{h}$$ VS u, $$u_{m}^{h}(0.5)-u(0.5)$$ VS m

solution
1. According the definition of constraint breaking solution(CBS), we can prove the LIBF can satisfy the CBS.

 * $$l_{m}^{i}(x)=\prod\limits_{j=1,j\ne i}^{m}{\frac{(x-{{x}_{j}})}{({{x}_{i}}-{{x}_{j}})}}$$

where,
 * $${{x}_{1}}=0,{{x}_{2}}=1/3,{{x}_{3}}=2/3,{{x}_{4}}=1$$.

At the boundary of $$\beta $$ , it can be solved that: $$\begin{align} & \beta =0, \\ & l_{m}^{1}(x=0)=1; \\ & l_{m}^{2}(x=0)=l_{m}^{3}(x=0)=l_{m}^{4}(x=0)=0 \\ \end{align}$$ So, these four basis function can satisfy the CBS. The procedure is the same for m=6, m=8 here. 2. Plot all the basis used for m=4, m=6, m=8, respectively. m=4 m=6 m=8 3. The exact solution is obtained by following steps. $$\frac{\partial}{\partial x}\left[(2+3x)\frac{\partial u}{\partial x}\right]+5x=0$$ Integrate the above equation, we obtain that: $$\left[(2+3x)\frac{\partial u}{\partial x}\right]+\frac{5x^{2}}{2}+C_1=0$$ Applying the natural boundary condition: $$\Gamma _{h}=\left\{0\right\}, h = 12$$ Then, we obtain that: $$\mathbf{C_1 = -12}$$ Then, we get: $$\left[(2+3x)\frac{\partial u}{\partial x}\right]+\frac{5x^{2}}{2}-12=0$$ Using WA to solve the equation, and applying the essential boundary condition yields the exact solution:

$$u=(-45{{x}^{2}}+60x-472\centerdot log(3x+2)+60)/108+4+(472\centerdot log(5)-75)/108$$

4. m=4, the exact solution and approximate solution using LIBF are compared in the following figure: The according matlab code is: m=6, the exact solution and approximate solution using LIBF are compared in the following figure: The according matlab code is:

m=8, the exact solution and approximate solution using LIBF are compared in the following figure: The according matlab code is: The error is decreased by increasing the nodes of basis function(LIBF), which can be clearly in the following figure: The according matlab code is:

This problem was solved by shengfeng yang

=Problem 5.7 Linear Lagrangian Element Basis Function with dataset G1DM1.0/D1b=

From the lecture slide Mtg30

Problem Statement
Similar to HW5.1/HW5.3, but using Linear Lagrangian Element Basis Function (LLEBF ) with uniform discretization (equidistant element nodes). The number of element ($$\displaystyle nel$$) is 4,6,8.

Find:
a) Explain how LLEBF are used as CBS, Constraint Breaking Solution

b) Plot all LLEBF used for $$\displaystyle nel=4$$

c) Use matlab to calculate

d) Plot figures showing $$\displaystyle {{u}^{h}}$$ vs $$\displaystyle u$$, $$\displaystyle \left[ u_{nel}^{h}(0.5)-u(0.5) \right]$$ vs number of element

a) Explain how LLEBF are used as CBS
In the dataset G1DM1.0/D1b, $$\displaystyle {{\Gamma }_{g}}=\left\{ \beta \right\}=\left\{ 0 \right\}$$. As to the Constraint Breaking Solution, it’s required that,

The LLEBF has similar property with Lagrangian Interpolation function in,

Hence, the family $$\displaystyle F=\left\{ {{b}_{0}}(.),{{b}_{1}}(.),{{b}_{2}}(.),\cdots \cdots ,{{b}_{n}}(.) \right\}$$ will satisfy the CBS requirement in that,

No need for constant basis function this case

Generally the family of basis function should include constant one. However there’s no need for LLEBF basis, since the summation $$\displaystyle \sum\limits_{0}^{n}{{{b}_{i}}(.)}$$ can form a constant. That’s property of LLEBF, see part b) below.

b) Plot all LLEBF used for the case of 4 elements
See the figures below

We also post the LLEBF used for case of 6 and 8 elements

c) Use matlab to calculate
The MatLab code is included in the collapsible box

Matlab code for the $$\displaystyle nel=4$$ case

Matlab code for the $$\displaystyle nel=6$$ case

Matlab code for the $$\displaystyle nel=8$$ case

The author of the code also invented a "truncate" function to help calculation:

Matlab Code for "truncate" function

d) Plot figures showing comparison between exact and numerical solution, convergence vs number of element
From HW5.1/HW5.3 we have known the exact solution is,

The comparison of $$\displaystyle {{u}^{h}}$$ vs $$\displaystyle u$$:

$$\displaystyle \left[ u_{nel}^{h}(0.5)-u(0.5) \right]$$ vs number of element:

We can see that the convergence becomes better when number of elements increase.

d) Plot figures showing comparison between exact and numerical solution, convergence vs number of element
From HW5.2/HW5.4 we have known the exact solution is,

The comparison of $$\displaystyle {{u}^{h}}$$ vs $$\displaystyle u$$:

$$\displaystyle \left[ u_{nel}^{h}(0.5)-u(0.5) \right]$$ vs number of element:

We can see that the convergence becomes better when number of elements increase.

title: User:Eml5526.s11.team4.yang/hw5


title: File:Hw5.4-m=4.jpg


title: File:Hw5.4-m=4-basis.jpg


title: File:Hw5.4-m=6.jpg


title: File:Hw5.4-m=6-basis.jpg


title: File:Hw5.4-m=8-basis.jpg


title: File:Hw5.4-m=8.jpg


title: File:Hw5.4-comp-err.jpg


text: File:QLEBF line.png

title: File:QLEBF line.png


text: File:Quad Lagrangian Basis functions.png


text: File:Eml5526.s11.team2.brenner.hw5.3.jpg


text: File:Eml5526.s11.team2.brenner.hw5.3.6nodes.jpg


text: File:Eml5526.s11.team2.brenner.hw5.3.8nodes.jpg


text: File:Hw5.7-cmp error.png

title: File:Hw5.7-cmp error.png


title: File:Hw5.7-m=4.png


title: File:Hw5.7-m=6.png


title: File:Hw5.7-m=8.png


text:
User:Egm6341.s11.team4.kurth/hw5

=Problem 5.5: Modifying MATLAB Code to Calculate the Composite Trapezoidal Rule More Efficiently=

Objective

 * 1) Modify the MATLAB code used in HW 2.4 to make calculating the composite trapezoidal rule more efficient, i.e., $$ \displaystyle T_0(2^j)=T_0(2^{(j-1)})+\cdots$$
 * 2) Create a Romberg table and compare results to those initially calculated in HW 2.4

Modified MATLAB Code and Development Process
Recall the formula for the composite trapezoidal rule

 * $$T_0(n)=\frac{(b-a)}{n}\left(\frac{1}{2}f(x_0)+f(x_1)+f(x_2)+\cdots+f(x_{n-1})+\frac{1}{2}f(x_n)\right)$$

where $$ h = \frac{(b-a)}{n},,\;\;x_0=a,$$ and $$\displaystyle x_n=b$$. In practice it is common to successively double n (typically a power of two to begin with), the number of panels used, until a satisfactory convergence is obtained, effectively halving each subinterval. The new set of nodes would include all of the previously used $${x_i, i=0,1,2,...,n}$$ as well as new nodes which bisect the previous panels. To increase computation efficiency, the values of the function evaluated at all the nodes during the previous computation will be reused to calculate successive estimates. This yields the following relationship

Note that the $$\frac{1}{2}$$ factor in the result of (5.1) is due to the halving of the step-size; $$ h_{new} = \frac{1}{2}h_{old}$$.

This concept was employed to modify a previous MATLAB code to use this technique to cut down on computational cost. The modified code is found below.

Romberg Table
The Romberg Table involves using previously calculated numerical estimates of an integral to produce higher order estimates to surprisingly high accuracy. Equation [[media:nm1.s11.mtg29.djvu| (3)p.29-5]] gives the relation

A MATLAB code was developed to take the output of the modified composite trapezoidal rule and create a Romberg table and is found below:

These codes were used to find estimates of the integral

The following MATLAB script performed the computations for this case:

Matlab Code:

Comments
The 2nd column, corresponding to the composite trapezoidal rule approximations match those previously calculated in 2.4 exactly. It it observed that for higher levels of the trapezoidal rule (higher k) that a much smaller number of panels is needed for far greater accuracy. This shows that the Romberg table is a powerful tool for acquiring accurate numerical estimations of integrals using minimal computational effort.

Solved by William Kurth.

=Problem 5.7: Pros and Cons of Various Numerical Integration Techniques=

Objective
Discuss the pros and cons of the following numerical integration methods:
 * 1) Taylor Series
 * 2) Composite Trapezoidal Rule
 * 3) Composite Simpson's Rule
 * 4) Romberg Table(Richardson Extrapolation)
 * 5) Corrected Trapezoidal Rule

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


text:
Portal talk:Spanish/Spanish stream

BlueSkies305 09:08, 23 March 2011 (UTC)BlueSkies305

title: Portal talk:Spanish/Spanish stream


text:
User:Eml5526.sp11.team1.adam/HW5

=Problem 5.1: Weak form with appropriate functions such as Polynomial, Fourier & Exponential until convergence.=

Given
General 1-D Model 1 Data set 1. The Partial Differential equation for this data set is,

Find
Solve the Weak Form of the above general model 1 data set 1 for the Functions such as

1)Polynomial Basis Function,

2)Fourier Basis Function &

3)Exponential Basis Function

to find out the convergence point.

1) Polynomial Basis Function
Solve the Differential Equation

in the interval (0,1) with the boundary conditions

The Weak Form of the given Differential Equation will be obtained as follows

So we need to integrate the above equation using integration by parts, after which we will get the following equation,

{| style="width:100%" border="0" cellpadding="5" cellspacing="0" $$\left(5.1.5\right)$$
 * style="width:10%" |
 * style="width:80%" |
 * style="width:10%" |
 * style="width:10%" |

We now have to select the appropriate Polynomial Basis Function which will follow the constraint breaking solution. so selecting

Clearly,

We need, So Choosing,

We now have, Using the essential Boundary condition we have, Therefore, In the identical way we can also write,

With the condition,

Therefore,

Now we need to calculate the derivative of $$U$$ & $$w$$, Therefore,

{| style="width:100%" border="0" cellpadding="5" cellspacing="0" $$\left(5.1.9\right)$$
 * style="width:10%" |
 * style="width:80%" |
 * style="width:10%" |
 * style="width:10%" |

Substituting this into the weak form of $$\left(5.1.9\right)$$, we get,

Cancelling $$c_{i}$$ and Organizing, we get, Simplifying the last equation we get,

We can obtain the value of $$\displaystyle a_{j}$$ from the above equation.

The analytical exact solution on the other hand is obtained as,

therefore,

The Values of p & r are obtained by using the following boundary conditions,

We get, Substituting in the above equation $$\left(5.1.11\right)$$ we get,

{| style="width:100%" border="0" cellpadding="5" cellspacing="0" $$\left(5.1.12\right)$$
 * style="width:10%" |
 * style="width:80%" |
 * style="width:10%" |
 * style="width:10%" |

To solve this system of equations we developed a MATLAB code. It was found out that with $$n=2$$ a good approximation is obtained. However with$$n=8$$ the error was smaller than $$10^{-6}$$. The MATLAB code is as follows.

Appendix
 Matlab Code:

The analytical and the actual function along with the Error between the two are plotted as below.

2) Fourier Basis Function
To follow the constraint Breaking Solution we choose,

We need to satisfy the essential boundary condition So we choose the basis Function such that So we can write, Now the trial solution will be as follows,

Now we need to solve for the remaining coefficients By constructing a Stiffness & Force Matrices as follows.

3) Exponential Basis Function
To follow the constraint Breaking solution we choose: Clearly, And as we need We choose, Eventually we have, From the essential boundary condition, we have, Therefore, In the identical way we have, Using the condition, Therefore Derivative of U & w will be,

Replacing in the weak form of $$\left(5.1.5\right)$$ we get,

Cancelling $$c_{i}$$ and then organizing we obtain, Performing the integration we obtain following equation,

The value of the $$a_{j}$$can be calculated from the above equation. we developed a MATLAB code to find the error between the Exact and the analytical solution. The MATLAB code is as follows.

Appendix
 Matlab Code:

The following two figures shows the basis Function and the Error between the Exact and the analytical solution. It can be observed that the Exponential basis Function converges Slowly than the Polynomial basis Function. For exponential the convergence was achieved with 18 terms whereas with polynomial, it was achieved with only 8 functions. Moreover the stiffness matrix obtained with exponential functions was bad-conditioned for a number of functions greater than 8

=Problem 5.2: ...=

Solution
=Problem 5.3: ...=

Solution
=Problem 5.4: ...=

Solution
=Problem 5.5: Calculix=

Given
Given the disk problem in homework four [| Working with Calculix]

Find

 * 5.1 Extract node information to include node numbers and coordinates, and element information to include element numbers and element nodes.
 * 5.2 Generate 3 meshes of same disk with triangular elements.
 * 5.3 Install ccx, run examples, write report for “dummies”

5.1 Node and Element information
A disk was created in Problem 4.7 in the previous homework. CGX was opened with this file, and the command “prnt se 1” was used to extract the information. This was copied to a file and is available at the following link http://en.wikiversity.org/wiki/File:Fe1.s11.team1.hw5.node.element.info.pdf

5.2 Three Meshes
Element type was changed using the command “elty all tr3”, “etly all tr3u” and “elty all tr6” and then using the “mesh all” command to mesh the disk. Refer to figures 5.1, 5.2 and 5.3

5.3 CCX
CCX was installed and opened. The manual spells out the program well; this was downloaded, but can be found at http://web.mit.edu/calculix_v2.0/CalculiX/ccx_2.0/doc/ccx/index.html. Examples were run, please see the following report for findings. http://en.wikiversity.org/wiki/File:CCX.pdf

Contributing Members
Posted by James Roark 01:59, 23 March 2011 (UTC)

=Problem 5.6: Quadratic Lagrangian Elemental Basis Function.=

Given
Consider any Quadratic Function.

Find
Plot of the Individual Lagrangian Quadratic Function and the combined plot of all the three Lagrangian Quadratic Functions.

Solution
For any Quadratic Function the Number of nodes are always 3. Therefore we can write, We will select the global node numbers as follows, We have according to the Lagrangian Interpolation Basis Function, Which is the Polynomial of degree (n-1)

So, Substituting the values from Equation $$\displaystyle (5.6.2)$$, we get,

In the Similar way,

&

We Developed a MATLAB code to plot all these Quadratic Lagrangian Basis Function. This very MATLAB code is as follows.

Appendix
 Matlab Code:

The plots of all the Individual Lagrangian Quadratic Function and the Combined plot of all the Three Functions are as follows.

=Problem 5.7: Linear Lagrangian Element Basis Function= Solving a ordinary differential equation using Linear Lagrangian Element Basis Function. From Mtg.30

Given
The partial differential equation is:

The boundary conditions are:

$$ u\left(0\right) = 4$$

$$\frac{du}{dx}\left(1\right) = -6 $$

Find

 * 1. Explain how Linear Lagrangian Element Basis Function (LLEBF) are used as Costrain Breaking Solution
 * 2. Plot all LLEBF
 * 3. Solve the system dividing the interval [0,1] in 3 divisions (m = 4) and continue dividing until convergence of 10^-6 of the error at x = 0.5
 * 4. Plot the exact and the aproximated solution
 * 5. Plot the error as a function of the number of points

Explain how LLEBF are used as CBS
A function $$b_{i}$$ is used as CBS if: $$b_{0}\left(\beta\right) = 1$$, and $$b_{i}\left(\beta\right) = 0; i = 1, 2, ...$$.

A funtion u(x) can be interpolated by using linear lagrange polynomial considering an element of two points (nodes): 1 and 2; the function will be given by:

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

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

Where $$u_{1}, u_{2}$$ are the nodal values of the funtion at the nodes 1 and 2 respectively.

The Lagrange Polynomial has the property of have the value of 1 at the respective node an 0 at the others; for example in equation (5.7.2), if an esential boundary condition is applied on node 1, whe have that $$N_{1} = b_{1} = 1$$ and $$N_{2} = b_{2} = 0$$ at the node 1 where the essential boundary condition is applied. In this case we only have two fuctions (b1 and b2); therefore this polynomial follows CBS. Symilar aproach can be used if the essential boundary condition is at node 2.

Plot all LLEBF used
Fig. 5.7.1 shows the two basis functions described in Eq. (5.7.2); in the figure, clearly can be seen that N1 is equall to 1 at node 1 and 0 at node 2, and N2 is equal to 0 at node 1 and equal to 1 at node 2. The plot are obviously straight lines because the functions are linear.

In the present problem we have to start working with 4 poins (3 elements) and increase the number of points until that the error at x = 0.5 is smaller than 10^-6. We are asked to plot all the basis functions used, but the basis funtion used are similar for all elements does not matter the size of the element. When the number of elements is increased, the lenght of the elements is decreased, but the basis fuction are the same as that shown in Fig. 5.7.1 for all the elements. Obviously a function is valid only inside its respective element

Solve the system dividing the interval [0,1] in 3 divisions (m = 4) and continue dividing until convergence of 10^-6 of the error at x = 0.5
Firs we obtain the weak form of the differential equation; to do that we multiply it for a weight funtion and integrate from 0 to 1:

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

$$\int_{0}^{1}w\left ( \frac{d}{dx} \left [ \left ( 2+3x \right ) \frac{du}{dx}\right ]+5x\right )dx=0 $$
 * $$\displaystyle (5.7.3) $$
 * }
 * }

Performing integration by parts in the first term we obtain:

As we do not know the derivative of u at x = 0, we chose w(0) = 0. The global stiffness matrix will be assembled from the stiffness matrix of each element. To get the element stiffness matrix we change the cordinate x to the cordinate $$\epsilon$$ which is equal to -1 at the node 1 and is equal to 1 at the node 2, as indicated in the Fig. 5.7.2

The new cordinate $$\epsilon$$ is given by:

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

$$\varepsilon = \frac{2\left ( x-x_{1} \right )}{x_{2}-x_{1}}-1=\frac{2\left ( x-x_{1} \right )}{l}-1 $$ In the last step, it was used the fact that $$x_{2}-x_{1} = l$$; the lengt of the element. Taking the derivative on both sides of equation (5.7.5) we have:
 * $$\displaystyle (5.7.5) $$
 * }
 * }

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

$$ d\varepsilon =\frac{2dx}{l}$$
 * $$\displaystyle (5.7.6) $$
 * }
 * }

Replacing the cordinate x in the lagrange polynomial we obtain:

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

$$u\left ( \varepsilon \right )=\begin{bmatrix} \frac{1-\varepsilon }{2} & \frac{1+\varepsilon }{2} \end{bmatrix}\begin{bmatrix} u_{1}\\

u_{2}\end{bmatrix}=\left [ N \right ]\begin{bmatrix} u_{1}\\

u_{2}\end{bmatrix} $$
 * $$\displaystyle (5.7.7) $$
 * }
 * }

In symilar way w can be written as:

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

$$\left [w\left(x\right)= N \right ]\begin{bmatrix} c_{1}\\

c_{2}\end{bmatrix}=\begin{bmatrix} c_{1} & c_{2} \end{bmatrix}\left [N \right ]^{T} $$ As in the weak form, we need the derivatives of u and w with respect to x, we calculathe those derivatives by using the chain rule:
 * $$\displaystyle (5.7.8) $$
 * }
 * }

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

$$\frac{du}{dx}=\frac{du}{d\varepsilon }\frac{d\varepsilon }{dx}=\frac{du}{d\varepsilon }\frac{2}{l} $$
 * $$\displaystyle (5.7.9) $$
 * }
 * }

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

$$\frac{dw}{dx}=\frac{dw}{d\varepsilon }\frac{d\varepsilon }{dx}=\frac{dw}{d\varepsilon }\frac{2}{l} $$ From Eq. (5.7.7) we can take the derivative with respect to \epsilon to obtain:
 * $$\displaystyle (5.7.10) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$\frac{du}{d\varepsilon }=\begin{bmatrix} -\frac{1}{2} & \frac{1}{2} \end{bmatrix}\begin{bmatrix} u_{1}\\

u_{2}\end{bmatrix}=\left [ B \right ]\begin{bmatrix} u_{1}\\

u_{2}\end{bmatrix} $$ In similar way the derivative of w is:
 * $$\displaystyle (5.7.11) $$
 * }
 * }

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

$$\frac{dw}{d\varepsilon }=\begin{bmatrix} c_{1} & c_{2} \end{bmatrix}\left [ B \right ]^{T} $$
 * $$\displaystyle (5.7.12) $$
 * }
 * }

Making the change of variables on the stiffness matrix and on the force vector we get:

To assemble the system we build a Matlab code (shown below) wich basically follow the algoritm described on Fig. 5.7.3. Inside the code, the integration was done by using the Gauss quadrature method. In that method, the integral of a funtion is given by:

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

$$ \int_{-1}^{1}f\left ( \varepsilon \right )d\varepsilon =\sum p_{i}f\left ( \varepsilon _{i} \right ) $$
 *
$$\displaystyle (5.7.15) $$
 * }
 * }

Where $$p_{i}$$ are weighting coefficients. As in Equation (5.7.14) we have a polynomial of order 2, we need at least 2 coefficients to exactly calculate the integral. In this case the integral is given by:

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

$$ \int_{-1}^{1}f\left ( \varepsilon \right )d\varepsilon =f\left ( \frac{1}{\sqrt{3}} \right )+f\left ( \frac{-1}{\sqrt{3}} \right ) $$
 *
$$\displaystyle (5.7.16) $$
 * }
 * }

Plot the exact and the aproximated solution
Fig. 5.7.4 and Fig. 5.7.5 shows the exact and aproximated solution obtained with 4 and 6 nodes respectively. It can be observed that even with 4 nodes the numerical solution aproximates well to the analytical.

To calculate the error, at x = 0.5 is the mid point of the domain, we obtain the value at x = 0.5 interpolating the two neighboring values. The interpolation was done by using the same Lagrange polynomial described before. We found that even though a good aproximation was obtained even with low number of elements (n = 4, Fig. 7.5.4) the convergence to a tolerance of 10^-6 was not obtained with a number of elements lower than 100. Therefore we decrease the tolerance to 10^-3. With this this tolerance the convergence still was achieve with a large number of elements (19) Fig.7.5.6 shows the error as a funtion of the number of elements.

Appendix
 Matlab Code:

=Problem 5.8: Linear Lagrangian Element Basis Function= Solving a ordinary differential equation using Linear Lagrangian Element Basis Function. From Mtg.30

Given
The partial differential equation is:

The boundary conditions are:

$$ u\left(1\right) = 4$$

$$\frac{du}{dx}\left(0\right) = -6 $$

Find

 * 1. Explain how Linear Lagrangian Element Basis Function (LLEBF) are used as Costrain Breaking Solution
 * 2. Plot all LLEBF
 * 3. Solve the system dividing the interval [0,1] in 3 divisions (m = 4) and continue dividing until convergence of 10^-6 of the error at x = 0.5
 * 4. Plot the exact and the aproximated solution
 * 5. Plot the error as a function of the number of points

Solution
This problem is similar to 5.7 but only change the boundary conditions, therefore, the weak form is the same exept that now we chose w(1)=0 and impose the boundary condition u(1) = 4. As the problem is similar, we use the same Matlab routine developed for porblem 5.7. As all the develpment is the same that in problem 5.7 here we only present the results. Fig. 5.8.1 and Fig. 5.8.2 show the analytical and numerical solution obtained with 4 and 6 nodes respectively; again, although good aproximation was obtained even with 4 nodes, no convergence was achieved even with 100 nodes; therefore, the tolerance was decreased to 10^-3. With this new tolerance, convergence was achieved with 8 nodes. Fig. 5.8.3 shows the error as funtion of the number of nodes.

=Please update contributing members table with appropriate info= =Contributing Members=

title: User:Eml5526.sp11.team1.adam/HW5


title: File:EML.5526.s11.tm7.HW5.3Pic1.1.jpg


text: File:EML.5526.s11.tm7.HW5.3Pic1.jpgEML.5526.s11.tm7.HW5.3Pic1.1.jpg


text: File:EML.5526.s11.tm7.HW5.3Pic1.jpgEML.5526.s11.tm7.HW5.3Pic2.jpg


text: File:EML.5526.s11.tm7.HW5.3Pic1.jpgEML.5526.s11.tm7.HW5.3Pic3.jpg


text: File:EML.5526.s11.tm7.HW5.3Pic1.jpgEML.5526.s11.tm7.HW5.3Pic4.jpg


text: File:Hw5388.jpgHw5366.jpgeml.5526.s11.tm7.Hw5388.jpg


text: File:Hw5366.jpgeml.5526.s11.tm7.Hw5366.jpg


title: File:Eml.5526.s11.tm7.Hw5344.jpg


title: File:Eml.5526.s11.tm7.Hw5322.jpg


title: File:Eml.5526.s11.tm7.Hw5382.jpg


title: File:Eml.5526.s11.tm7.Hw5381.jpg


title: File:Eml.5526.s11.tm7.Hw536.jpg


title: File:Eml.5526.s11.tm7.Hw534.jpg


title: File:Eml.5526.s11.tm7.Hw532.jpg


title: File:Eml.5526.s11.tm7.Errorplot.jpg


text:
User:Egm5526.s11.team2.sandhu/newpage

new page home work 5 =Problem Weak form solution using LLEIF= Solve the following discrete weak form with Linear Lagrange Element Interpolation Basis.

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

$$  \displaystyle \sum {{c_i}\left[ {\sum {{{\tilde K}_{ij}}{d_j} - {{\tilde F}_i}} } \right]} = 0 $$     (3.7.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Where

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

$$  \displaystyle {{\tilde K}_{ij}} = \int\limits_0^1 {{{b'}_i}(2 + 3x){{b'}_j}dx} $$     (3.7.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle {F_i} = {b_i}\left( 1 \right)\left( {12} \right) + \int_0^1 {{b_i}\left( x \right)5x} dx $$ (3.7.3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Such that

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

$$  \displaystyle {u^h}\left( 0 \right) = \sum {{d_j}{b_j}\left( 0 \right) = 4} $$     (3.7.4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle {w^h}\left( 0 \right) = \sum {{c_i}{b_i}\left( 0 \right) = 0} $$     (3.7.5)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

d. Plot $${u^h}$$ vs $${u}$$
Plot $${u^h}\left( {0.5} \right) - u\left( {0.5} \right)$$ vs n (total no. of dof's)

a. Using LLEBF in constraint breaking solution(CBS)
We have following constraint to be satisfied to use weak form

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

$$  \displaystyle {w^h}\left( 0 \right) = \sum {{c_i}{b_i}\left( 0 \right) = 0} $$     (3.7.6)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

At node 1 (i.e. at x=0) all the basis function are zero except 1.

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

$$  \displaystyle {w^h} = {c_1}{b_1} + {c_2}{b_2} + {c_3}{b_3} + {c_4}{b_4} + \cdots $$     (3.7.7)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle {w^h} = {c_1}(1) + {c_2}(0) + {c_3}(0) + {c_4}(0) + \cdots $$     (3.7.8)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle {w^h} = {c_1} $$     (3.7.9)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle {w^h}\left( 0 \right) = 0 $$     (3.7.10)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle {c_1} = 0 $$     (3.7.11)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

To satisfy homogeneous boundary condition while using Linear Lagrange Element Interpolation Functions, we just have to select our $$w^h$$ such that $${c_1} = 0$$.

c Formulation of K and F on element level
We can divide the total domain into elemental domain.

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

$$  \displaystyle \Omega = \bigcup\limits_{e = 1}^{nel} $$     (3.7.12)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Values of K and F will be summation of K and F over the elements
 * {| style="width:100%" border="0"

$$  \displaystyle {{\tilde K}_{ij}} = \sum\limits_{e = 1}^{nel} {K_{ij}^e} $$     (3.7.13)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle K_{ij}^e = \int\limits_ {{{b'}^e}_i(2 + 3x){{b'}^e}_jdx} $$     (3.7.14)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle {{\tilde F}_i} = \sum\limits_{e = 1}^{nel} {F_i^e} $$     (3.7.15)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle {F_i}^e = \int_ {b_i^e5xdx} $$     (3.7.16)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

There will be two basis function for each element. Generalised values of these basis function will be as follow

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

$$  \displaystyle {b_e} = \frac $$     (3.7.17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle {b_{e + 1}} = \frac $$     (3.7.18)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

d. Solution using Matlab
Plot of exact solution and approximate solution with 2 elements

Plot of exact solution and approximate solution with 4 elements

Plot of exact solution and approximate solution with 6 elements

Plot of exact solution and approximate solution with 8 elements

Plot of error vs dof's

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

$$  \displaystyle err = u\left( {0.5} \right) - {u^h}\left( {0.5} \right) $$     (3.7.19)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle dof's = nel + 1 $$     (3.7.20)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

where nel= number of elements.

for 16 elements

dof's=16+1=17

comment
Solution is converging with increasing no. of elements. For n=16 error is less than 0.0001.

title: User:Egm5526.s11.team2.sandhu/newpage


text: File:Fe1.s11.tm7.HW5.PB5.2 Fourier.pdf

title: File:Fe1.s11.tm7.HW5.PB5.2 Fourier.pdf


title: File:Fe1.s11.tm7.HW5.1.exp1.jpg


title: File:Fe1.s11.tm7.HW5.1.exp2.jpg


title: File:Fe1.s11.tm7.HW5.1.exp3.jpg


title: File:Fe1.s11.tm7.HW5.1.exp10.jpg


title: File:Fe1.s11.tm7.HW5.1.exp20.jpg


text: File:Fe1.s11.tm7.HW5.PB5.1 Fourier.pdf

title: File:Fe1.s11.tm7.HW5.PB5.1 Fourier.pdf


text: File:Largetable.png

title: File:Largetable.png


text: File:Smalltable.png

title: File:Smalltable.png


title: File:Mesh1.png


title: File:Mesh2.png


title: File:Mesh3.png


text:
User:Eml5526.s11.team4.rhee/HW5

= Problem 5.2 - G1DM1.0/D1 using WF =

Problem Statement
Solve G1DM1.0/D1 using WF with appropriate basis function (Poly, Fourier, Exp), until convergence of $$u^h(0.5)$$ to $$O(10^{-6})$$.

Boundary Conditions
The boundary conditions are

Solution
Starting with the differential equation 3 from page 9-2 in the notes

Applying the boundary condition, $$-2du/dx(x=0)=12$$

Applying the boundary condition $$u(1)=4$$

Plot

Using Discrete WF, find $$ \left\{\mathbf{d_j}\right\} $$ st $$u^h(\beta)=\sum_{j}\mathbf{d_j}(t)\mathbf{b_j}(\beta)=g \quad $$ and

where

 * $$\mathbf{K_{ij}}=\int_{0}^{1}\mathbf{b_i}'a_2\mathbf{b_j}'dx$$
 * $$\mathbf{F_i}=\mathbf{b_i(\alpha)}h+\int_{0}^{1}\mathbf{b_i}fdx$$
 * }
 * $$\mathbf{F_i}=\mathbf{b_i(\alpha)}h+\int_{0}^{1}\mathbf{b_i}fdx$$
 * }

Polynomial basis function

 * $$\mathfrak{F}_{p}=\left\{(x-1)^i,i=0,1,2,...,n\right\}$$

When n = 8, $$ Error \approx 0.1\times 10^{-6} $$

 * $$K_{ij}\approx \begin{bmatrix}

3.5 & -3 & 2.75 & -2.6 & 2.5 & -2.42 & 2.37 & -2.33\\ -3 & 3.66 & -3.9 & 4 & -4.04 & 4.07 & -4.08 & 4.08\\ 2.75 & -3.9 & 4.5 & -4.85 & 5.08 & -5.25 & 5.36 & -5.45\\ -2.6 & 4 & -4.85 & 5.42 & -5.83 & 6.13 & -6.36 & 6.54\\ 2.5 & -4.04 & 5.08 & -5.83 & 6.38 & -6.81 & 7.15 & -7.43\\ -2.42 & 4.07 & -5.25 & 6.13 & -6.81 & 7.36 & -7.80 & 8.17\\ 2.37 & -4.08 & 5.36 & -6.36 & 7.15 & -7.8 & 8.34 & -8.8\\ -2.33 & 4.08 & -5.45 & 6.54 & -7.43 & 8.17 & -8.8 & 9.33 \end{bmatrix}, \quad for \quad i,j=1,2,...,8 $$

 * $$F_{i}\approx \begin{bmatrix}

-12.83\\ 12.41\\ -12.25\\ 12.16\\ -12.11\\ 12.08\\ -12.06\\ 12.05 \end{bmatrix}

\qquad d_{j}\approx\begin{bmatrix} -2.89\\ 0.37\\ -0.27\\ 0.32\\ 0.43\\ 0.81\\ 0.61\\ 0.22 \end{bmatrix}, \quad d_0=4$$

Therefore,

Plot
The following figure shows $$ u^h(x)$$ The following figure shows the error vs n.

 * $$ Error\approx \begin{bmatrix}

-0.239809504267762\\ -0.008831243398196\\ 0.007978831249932\\ 0.000140951761964\\ -0.000263703709527\\ -0.000003409109291\\ 0.000009890451250\\ 0.000000100788939 \end{bmatrix}$$

Fourier basis function

 * $$\mathfrak{F}_{f}=\left\{1, cos\frac{(2j-1)\pi x}{2}, sin(k\pi x) ,j,k=1,2,...\right\}$$

Likewise, the following figure shows the error vs n.

When n=9, the error is $$0.000004240676477$$.

The following figure is $$ u^h(x) $$

Exponential basis function

 * $$\mathfrak{F}_{e}=\left\{1, e^{i(x-1)}-1 ,i=1,2,...\right\}$$

When n = 12, $$ Error \approx 0.63\times 10^{-6} $$ and that is,

 * $$ Error\approx \begin{bmatrix}

-0.297066776970831\\ -0.114733443435346\\ 0.0165040777664232\\ 0.012641225491135\\ 0.000082171723276\\ -0.001243314977745\\ -0.000190989291357\\ 0.000100445865361\\ 0.000035683793041\\ -0.000005235050710\\ -0.000004723596919\\ -0.000000063082541 \end{bmatrix}$$

Plot
The following figure shows the error vs n. The following figure shows $$ u^h(x)$$

I will show 4 figures when n=1,2,3,4 to see how $$ u^h(x)$$ change.

title: User:Eml5526.s11.team4.rhee/HW5


text: File:Ccxintro.png

title: File:Ccxintro.png


title: File:Hw5q7llebfplots.png


text:
User:Eml5526.s11.team04.premchand/HW5

= Problem 5.1 =

Problem Statement
Solve G1DM1.0/D1b using WF with appropriate basis function (Poly, Fourier, Exp), until convergence of $$u^h(0.5)$$ to $$O(10^{-6})$$.

Boundary Conditions
The boundary conditions are

Solution
On solving the given PDE with the boundary conditions we get ,the exact equation for u(x) is as follows

Polynomial
The polynomial basis function is given by $$\mathfrak{F}_{p}=\left\{x^{j},j=0,1,...,n\right\}$$. For $$\left\{j=0,1,2,3\right\}$$ the basis functions $$\left\{b_{j}\right\}$$ are as follows: The constraint breaking solution$$\left\{\bar{b}_j\right\}$$ for the polynomial basis function is shown below: In the specific case where $$\left\{\beta=0\right\}$$ the CBS becomes:

The basis functions$$\left\{\bar{b}_j\right\}$$ satisfy the CBS requirements since$$\left\{\bar{b}_0(\beta)=constant\neq0\right\}$$ and $$\left\{\bar{b}_j(\beta)=0, for j=1,2,...,n\right\}$$.

Matlab code for polynomial function
The following figure shows the original polynomial basis functions and the CBS basis functions for $$\left\{j=1,2,3\right\}$$,We can see that convergence occurs at an n value of 8

The following figure shows the error vs n

Exponential
The exponential basis function is given by So, the basis functions $$\left\{b_j\right\}$$ are as follows: The corresponding basis functions $$\left\{\bar{b}_j\right\}$$ satisfying CBS are shown below: The basis functions$$\left\{\bar{b}_j\right\}$$ satisfy the CBS requirements

 * $$\left\{\bar{b}_0(\beta)=1\neq0\right\}$$

 * $$\left\{\bar{b}_{j}(\beta)=0, for j=1,2,...\right\}$$.

In the specific case where $$\left\{\beta=0\right\}$$ the CBS becomes:

Matlab code for exponential function
The following figure shows the original polynomial basis functions and the CBS basis functions for $$\left\{j=1,2,3\right\}$$,We can see that convergence occurs at an n value of 12

The following figure shows the error vs n

title: User:Eml5526.s11.team04.premchand/HW5


title: File:5-3aa.gif


title: File:5-3b.png


title: File:5-3bb.png