text:
User talk:GerryWihardja

 Hello GerryWihardja, 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. 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.

 * 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.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Juan de Vojníkov 08:14, 6 February 2009 (UTC)

title: User talk:GerryWihardja


text:
User:Mhladnik

Miran Hladnik

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

title: User:Mhladnik


text: File:Logofellow.gif

title: File:Logofellow.gif


text:
Wikipedia and the 2008 US elections/Sarah Palin/Editor survey

The text below is a survey for the editors of the Sarah Palin article. Please don't feel the need to answer any questions you're not comfortable answering.

First contact with the article
Discuss your first contact with the article, for example: When did you first read the Sarah Palin Article? Did you think it was a good article when you first read it? When and why did you first edit the article? Do you think the article has improved since you first read it?

Background
Discuss your background knowledge of Sarah Palin for example: Were you familiar with Sarah Palin before John McCain's announcement on August 28, 2008? Do you have any academic background in politics or journalism? When you were editing, were you mostly adding material that you already knew, or were you actively researching and adding to the article as you found more information?

Politics
Discuss your political background, for example: Were you eligible to vote in this election? Did you vote? Are you registered to one of the parties? How did you vote? Did the Sarah Palin article or your involvement in the article help you decide how to vote?

Wikipedia
Discuss your views on the importance of Wikipedia to the election: Do you think the article played an important role in helping people decide who to vote for?

Editing habits
Discuss your editing habits: When did you first start to edit Wikipedia? Do you mostly work on politics-related articles?

Community on the Palin article
Discuss your experience as an editor of this article: Did you find that the community of editors working on this article were able to work well together? Did you ever get stressed? Did you develop a camaraderie with your fellow editors?


title: File:PEA2diagram.svg


text:
User talk:Egm6322.s09.Three/Report2

Gabriel, I have added in the proof for the linearity of differential operator after the definition of a linear operator in the beginning.

title: User talk:Egm6322.s09.Three/Report2


text:
User:Swift

mw:Extension:Quiz

title: User:Swift


title: File:Shepseskaf001.png


title: File:Khasekhemwy001.png


title: File:Netjerikhet001.png


title: File:Djoser001.png


title: File:Khasekhemwy-Seth-Horus001.png


text: File:Shepseskaf-mastaba.png

title: File:Shepseskaf-mastaba.png


title: File:Menkaure-pyramid001.png


title: File:Khafre-pyramid001.png


title: File:Djedefre-pyramid001.png


text: Swedish Language/Geography Terminology

title: Swedish Language/Geography Terminology


text:
User:Mitchaguilar

hi my name is mitchell and I'm currently attending the art institute of seattle for my project im going to make a fire poi video.

my paper on the olympic sculpture park, and the piece i chose is called Wake.

I went to the art sculpture park and I was drawn to many of the art pieces. But on that didn’t really catch my eye was one called wake. When I looked at it, it just looked like bit sheets of metal. I tried to be open-minded about it but it really wasn’t working for me. So I went home and looked up the artist name. The piece was done by Serra it consist of two huge sheets of curved steel that are mirror images of each other put together to form what Serra calls “Ws” or wiggles. After going to the park and doing some research, them re looking at the art. I have experienced a lot more insight and meaning on it.

Sometimes you look to close at the art instead of the gaps and space around it, that creates the movement. Instead of just looking at the steal look at the gaps and the air in the middle. I saw shapes and flow. Looking at other pieces that he has done were even more amazing. I’m glad that I took a sculpture that I didn’t find to impressing and researched it because now it has made me want to look at more artwork don’t by this artist and see in person. At first when I looked at this piece I was kind of distraught and confused on what the artist was trying to get across and explain. Once I did the research and going back to the park I felt the sculpture came to life.

title: User:Mitchaguilar


text:
User talk:Raghumadambath

i am starting

title: User talk:Raghumadambath


text:
Help talk:The original tour for newcomers/1

Note
It is normal and appropriate for newcomers to change this page such as. -- Jtneill - Talk - c 10:33, 7 February 2009 (UTC)

Radial Compressor Simulation
Radial Compressor simulation in CFD takes a ling time, as long as 2 weeks, as there are some imherent difficulties in the flow physics of the compressor. I was thinking if there was an alternate way to reduce this time, by some way, it could help most of the simulations finish fast..

Gender based violence
What ignites spark of evil and negativity? Velogen (discuss • contribs) 12:45, 13 November 2022 (UTC)


text:
User:Egm6322.s09/Report table/Best of report R2

 Under construction; not final (but close to being stable after a couple of weeks). The intention here is to document the best features in any report for the readers (including you); if you see any excellent features in any of the reports (including your team's) that I may have missed noticing, don't hesitate to let me know. I don't have time to read all reports in detail. Due to time constraint, I would only selectively looked at a few important features. I also added some annotations (e.g., related to errors, improvements, etc.) where I looked through for the benefit of everyone. Egm6322.s09 12:18, 7 February 2009 (UTC)

Latest archived version (possibly with comments): Team Bit Team Mafia Team Three

 * classnotes (including simple HW problems), graded over 100%

 * 2nd-order quasilinear PDEs
 * examples
 * verification of linearity

 * 2nd-order linear PDEs
 * matrix-operator form
 * constant coefficients
 * varying coefficients
 * equivalence between two forms: regular form and matrix form
 * HW: Condition for equality of mixed derivatives $$\displaystyle u_{xy} = u_{yx}$$. Best solution: Team Bit simpler to understand; Team Mafia the proof is a bit more complicated to read, but essentially the same. Note: Team Three good starting point, which coincides with the proofs in the other two reports, but need to use mean-value-theorem to conclude, and can use different step size along $$\displaystyle x$$ and $$\displaystyle y$$ to be more general as in the proofs in the other two reports; appears to do this proof on their own; good.
 * linearity of diffusion operator
 * linearity of derivative. Best solution: Team Mafia. Note: Team Bit Team Three gave a statement, but not a proof.
 * linearity of gradient operator
 * linearity of divergence operator
 * nonlinear transformation of coordinates
 * matrix-operator form in new coordinates

 * linearity, 2 definitions
 * definition 1 in class
 * definition 2 in Kolmogorov & Fomin, Selvadurai
 * additivity
 * homogeneity
 * examples of linear maps in higher dimensions, matrices, homogeneity
 * transformation of coordinates, rotation of coordinate system. Best solution: Team Bit see better explanation in class.  Note: Team Mafia only display the rotation matrix, and referred to a book, but not derived it.  Team Three did not do this problem.
 * affine mappings, not linear; example, rotation and translation
 * equivalence between two definitions, proof
 * sufficient condition
 * necessary condition


text: Template:Nb

title: Template:Nb


title: Template:Nb2


text:
Template:Google

Soft redirect: Help:Google

title: Template:Google


text:
Template talk:Google

Sorry, James, but what is the use of this template? It doesn't simplify typing and it creates an extra layer over the actual wiki-markup. Hillgentleman | //\\ |Talk 15:05, 7 February 2009 (UTC)
 * Sorry, its probably no use, but I'd been having problems remembering the wiki-markup and kept searching for it. So, maybe we could create a help page instead? -- Jtneill - Talk - c 22:47, 7 February 2009 (UTC)
 * FYI, have abandoned this template and transferred relevant material to Help:Google. -- Jtneill - Talk - c 02:05, 8 February 2009 (UTC)
 * No problems. google: is just one of the many interwiki maps. Hillgentleman | //\\ |Talk 17:03, 8 February 2009 (UTC)

title: Template talk:Google


text: Category:Human resource management

title: Category:Human resource management


text:
Nonkilling Human Biology/Human nature

If members of all human cultures are found to share a set of behaviors then do those "universal" behaviors reveal genetic predispositions that are part of human nature? Is it a part of human nature that humans have a genetic predisposition towards some specific cultural elements? For example, is a genetic predisposition towards human language behavior part of human nature? Do humans have a genetic predisposition towards "territorial expansion and defense" ?

Do cultures based on a hunting and gathering lifestyle such as that of the ǃKung people provide evidence that humans do not have a genetic predisposition towards "territorial expansion and defense" ?

How many hunting and gathering people have been studied?

Paleodemography
Is it possible to measure and compare violent death rates in ancient human populations that lived either before or after the Neolithic Revolution? . A recent article by Samuel Bowles says that about 14% of deaths in 8 ethnographically studied hunter-gatherer cultures and 15 archaeologically studied hunter-gatherers cultures were apparently due to coalitions of members of a group inflicting bodily harm on one or more members of another group, suggesting that warfare is frequent and lethal in hunter-gatherer cultures.

Genetic predisposition
Much research onto human genetic predispositions receives funding because it is medically-oriented research that explores susceptibility to disease. Understanding of genetic predispositions to disease has been possible for many genes that have important functional roles in fundamental metabolic processes. For example, many gene variants have been linked to hereditary (inheritable) forms of cancer. Many well-studied examples of genetic predisposition to cancer are "high penetrance" examples that concern disease phenotypes that can be causally linked to gene mutations without a need to examine modulating effects of the patient's behavior or complex gene-environment interactions by which a person's cultural environment can significantly modify disease risk. However, some types of cancer show dramatic variations in incidence through time and between different cultures. The incidence of some forms of cancer is greatly increased by identifiable environmental factors and cultural practices. For example, the manufacture, advertising and sale of addictive tobacco-containing products can greatly increase the rate of smoking and the incidence of cancer. Does it make sense to say that humans have a genetic predisposition to nicotine addiction which has led to tobacco use in many cultures? Similarly, do some some humans have a genetic predisposition to alcohol addiction ? Can the genetics of violence be studied in the same way that other human behaviors (such as addiction-related behaviors) have been investigated ?

Evolution
If humans are part of a clade that icludes other member species that use violence in ways similar to humans, is that evidence for genetic predisposition towards human patterns of violence? For example, chimps have been seen killing members of other chimp groups in a way that is similar to lethal intergroup violence displayed by humans.

Related books

 * Shadows of Forgotten Ancestors, by Carl Sagan.
 * The Cambridge Encyclopedia of Hunters and Gatherers. See the introduction; Introduction: foragers and others by Richard B. Lee and Richard Daly.

title: Nonkilling Human Biology/Human nature


text:
User:BowmanL

Welcome to Life Skills for Lifelong Learners
The twenty-first century is one like no other in the history of the world. Communication, transportation, and commerce have created an ever expanding web of human interaction and interdependence. To flourish in this world, students need to learn to understand themselves more deeply and the world around them more thoroughly. This course is designed to help students do just that.

Begin this course by clicking on Unit 1 below.

 Table of Contents

Unit 1: Defining Myself
 * Self Discovery, Self-Esteem, Values, and Role Models

Unit 2: My Body, My Health
 * Smart Eating, Healthy Living, Stress and Conflict Management

Unit 3: My Place in the World
 * Family Ties, Citizenship, Global Communities

Unit 4: What The Future Holds
 * Envisioning the Future, Preparing for Adulthood

title: User:BowmanL


text:
Getting to Know You

This class is currently structured as a survey of basic desktop and internet browser-based software applications such as word processing tools, spreadsheets, presentations, and multimedia tools. I would like to explore context specific problems and issues. That will have to come from you.

Currently I know you as a class roster list, which is insufficient for our purposes. Please look at the course goals section of the syllabus and create a blog post concerning some of the terms that are in bold, as they relate to your current (or proposed) professional situation. This activity is a chance for you to begin to articulate context specific issues you may reflect upon during the semester. These items might complement other coursework. What can you tell the class about your personal inquiry interests, problems of personal relevance, issues of personal and/or social importance?

Microsoft Photo Story 3 This assignment is for Windows users only. Mac users please email me for further instructions. How does one create a digital movie for the Internet without a camera? You can now create slide shows using your digital photos. How much is this program? FREE. We will use this program later in the presentation module to teach a lesson, but for now the assignment is to tell your story using pictures and narration using Microsoft Photo Story. "Digital Storytelling is the modern expression of the ancient art of storytelling. Digital stories derive their power by weaving images, music, narrative and voice together, thereby giving deep dimension and vivid color to characters, situations, experiences, and insights. Tell your story now digitally." - Leslie Rule, Center for Digital Storytelling

This activity you will explore using Photo Story to tell your story. Have fun please.

Download and install Microsoft Photo Story and create a project, based upon still images, that will serve as a self introduction to the class. Add voice narration over your project. You have to have a microphone for this activity. Export your project as a movie. The My Video folder is a good option for storing your work on your computer. This will change your file from a Photo Story project which have a ( .wp3 ) file extension to a Windows Media File (.wmv ) file extension. View your exported project (movie) by double clicking on it. It should open in Windows Media Player. Publish your movie online. Huh? How? Where? Register for a You Tube or Teacher Tube account. Both will allow you to upload video and publish it where we can see it. Here is a YouTube exampleand an another example that is on the student's ftp site.

Knowing the Group Does anyone have any experience in online document sharing? Buckle up. You are about to. The big question of course is "How could I use this technology in my class"?

Things to consider. The document in question is a spreadsheet. The document is published as a web page. The document may be shared with people. This is the same technology as the word processor used in the resume activity. I will invite you to edit the document. You fill in the blanks. Distributed work! Less typing for me! Get it?

title: Getting to Know You


text:
Online graphics

 Online Graphic Tools

Your job is to create two online documents that contains some charts. There numerous online resources for this activity including [Crappy Graphics] and [Kid's Zone.] Recall you may create the graphics online, download them to your desktop (or the location of your choice) and then integrate them into your online documents.

Chart 1 should compare the 4th grade [National Assessment of Educational Progress (NAEP)] math scores for West Virginia compared to the national average for years 2000 through 2007. Write a brief statement of what you can conclude about West Virginia's performance.

Chart 2 should compare the number of students currently on an IEP for West Virginia and the next 3 poorest states based on per capita income. The source is a little dated but the trend is consistent. Write a brief statement of what you can conclude about West Virginia's student population in this category relative to its per capita income neighbors.

Intended Lesson Outcomes.

Student can create, integrate, publish, interpret, and discuss a graphic images.

title: Online graphics


text:
Help:Google

Examples of links which create a google search for a specific term:
 * search term - search term
 * Search google for "global financial crisis"+2009 - Search google for "global financial crisis"+2009

Note:
 * Use + instead of a space between the search terms

title: Help:Google


text:
Help:Firefox

Tips and help for using Wikiversity together with the Firefox web browser. Feel free to add good ideas or ask questions.

Quick search WV using FF shortucts

 * Go to Bookmarks - Organize Bookmarks (or CTRL-SHIFT-B) - R.click Bookmarks Toolbar or Bookmarks Menu to add a new bookmark. Suggest using:
 * Name: Wikiversity Address Bar Search
 * Location: http://en.wikiversity.org/wiki/Special:Search/%s
 * Keyword: v
 * Then, type v anything (e.g., help) in your address bar to search WV.

title: Help:Firefox


text:
User:Augur~enwikiversity

Hello hello, I'm an Advanced Diploma of Computer Systems Engineering graduate, with a perpetual need to correct english-based errors (as I'm sure some of you have found out from the edit histories from some of the pages I've read). I of course make these mistakes as well, however when I see them I feel I have to correct them. I blame my low grades in english from high school =).

I'm a perpetual learner, if there's something I don't know I strive to learn it, and thoroughly understand it (from my own perspective of course, as everyone understands the same concept differently).

Only recently found out about Wikiversity, and am exceptionally excited about the possibilities. I always want to do more courses and this is a way I can do them without having to worry about the bank balance and a good relationship with a disgruntled teacher. I have high hopes for the content here, and will add to (and correct spelling/grammar) wherever I can to help the site.

Should I make a gross error, please let me know. I don't mind being wrong and being corrected, but I like to know where I was wrong and why so as to improve myself and learn from my mistakes.

Enjoy yourself,

Augur.

title: User:Augur~enwikiversity


text:
User talk:Mhladnik

 Hello Mhladnik, 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. 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.

 * 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.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --mikeu talk 02:17, 8 February 2009 (UTC)

betawikiversity
Are you aware that that wikiversity has a multilingual project for non-english works? Visit http://beta.wikiversity.org for more information. --mikeu talk 02:20, 8 February 2009 (UTC)

 * Thanks for the information. I've moved Uvod_v_študij_slovenske_književnosti to http://beta.wikiversity.org/wiki/Uvod_v_%C5%A1tudij_slovenske_knji%C5%BEevnosti. Could you please delete the former page? --Hladnikm 08:52, 8 February 2009 (UTC)

 * So I have just deleted that.--Juan de Vojníkov 08:49, 15 March 2009 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Mhladnik. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Mhladnik~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:38, 17 March 2015 (UTC)

title: User talk:Mhladnik


text:
Interactive whiteboard

Smartboard tips

 * F11 for full-screen toggle

Optimising viewability (Win XP)

 * Use large icons
 * Screen resolution
 * Right click on a blank area of the desktop
 * Left click on Properties
 * Left click on the Settings tab.
 * Drag the Screen Resolution slider to 1280 by 768.
 * Click Apply and OK.
 * Use magnifier

Free Online / Web Whiteboards
Online web-based whiteboards support collaborative sharing of content.
 * Padlet
 * PixiClip
 * Stoodle
 * Web Whiteboard

title: Interactive whiteboard


text:
User talk:MostEarly

/kɒpiːˈɛdɪtə/ 05:39, 8 February 2009 (UTC)

title: User talk:MostEarly


text:
User:MostEarly

MostEarly=/kɒpiːˈɛdɪtə/

Please, call me Copy, MostĒ (or Mostē), Copy-Ed, or Amanacer (Spanish for 'to dawn' or when used with a definite article, 'the dawn', a reference to both my linguistic studies and my account's name.)

title: User:MostEarly


text:
Questionnaire

Surveys and questionnaires are used to collect qualitative and quantitative information about from a sample. Surveys of human populations and institutions are common in political polling and government, health, social science and marketing research. A survey may focus on opinions or factual information depending on its purpose, and many surveys involve administering questions to individuals. Surveys can be implemented via interview (in-person or via telephone) or self-report questionnaire (in-person, mail, email, web).

title: Questionnaire


text: Peer review

title: Peer review


text:
Talk:Survey design

Feel free to make suggestions or ask questions about survey design.

title: Talk:Survey design


text:
Swedish Language/Participles

Example sentences
Please translate into English  Maten är uppäten. De färdiga bilarna står uppställda utanför fabriken. Hon pratar flytande svenska. När vi hade åkt upptäckte vi att vi hade glömt att ta med oss någonting att dricka. Jag har kommit på någonting nytt!

title: Swedish Language/Participles


text:
User talk:Guhasutirtha

I am a 17 year old boy living in India. i am trapped in an extreme circumstanstial problem in which my education has been terribly ordealed. My country provides little oppurtunities and i have no option other than moving to a forgein land for further studies. I am in dire need. I need sponsorship for my education in the U.S./U.K. or any other country. I am starving for help. I am not financially not sound enough to pay the education costs. i have posted this comment to appeal for help.My email id is guhasutirtha@gmail.com  I request wikipedia to please help me in this regard.

title: User talk:Guhasutirtha


text:
User:Egm6322.s09.mafia/HW3

 My second wave of comments is color coded in green. Egm6322.s09 00:31, 2 April 2009 (UTC)

 See my comments below. Egm6936.s09 02:16, 21 February 2009 (UTC)

After you made a correction for a section with a comment box, you want to put a comment in that same comment box on what you did. Egm6936.s09 11:44, 26 February 2009 (UTC)

 NOTE: This page is being posted at roughly 19 UTC because of problems observed on the Wikiversity Domain. Anticipate revisions prior to 22 UTC, as well as possibly afterwards, if all edits are not possible due to the failures of the Wikiversity domain. --Egm6322.s09.lapetina 19:12, 20 February 2009 (UTC)

A few formatting and citation edits were made. No further edits are anticipated.

--Egm6322.s09.lapetina 20:11, 20 February 2009 (UTC)

Many edits have been made throughout. Editors were instructed to mention actions within comment boxes as indicated.

--Egm6322.s09.lapetina 21:05, 6 March 2009 (UTC)

=Transformations of Coordinates=

Linear Transformation of Coordinates
--Egm6322.s09.xyz 16:13, 4 April 2009 (UTC)

Example of Jacobian for Linear Transformation
Example: An example of a linear transformation is rotation in 2D. This can be expressed as:

$$

\begin{Bmatrix} \bar{x_1} \\ \bar{x_2} \end{Bmatrix}

=

\bar{R}

\begin{Bmatrix} x_1 \\ x_2 \end{Bmatrix}

$$

The transpose of the Jacobian matrix for this example is given as follows:

$$

\bar{J}^T

=

\begin{bmatrix} \frac{\partial \bar{x_1}} {\partial {x_1}} & \frac{\partial \bar{x_2}} {\partial {x_1}}\\ \frac{\partial \bar{x_1}} {\partial {x_2}} & \frac{\partial \bar{x_2}} {\partial {x_2}} \end{bmatrix}

=

\begin{bmatrix} R_{11} & R_{21}\\ R_{12} & R_{22} \end{bmatrix}

=

\bar{R}^T

$$

Nonlinear Transformation of Coordinates
--Egm6322.s09.xyz 16:14, 4 April 2009 (UTC)

Expression of general form in matrix notation using the Jacobian Matrix
Recall the general form of a 2nd order PDE that is linear in all orders expressed in matrix notation (Eqn. 2, pg. 8-1):

$$

\left \lfloor \partial_x \; \partial_y \right \rfloor \begin{bmatrix} a & b\\ b & c \end{bmatrix}

\begin{Bmatrix} \partial_x u \\ \partial_y u \end{Bmatrix} +

\left \lfloor d \; e \right \rfloor

\begin{Bmatrix} \partial_x u \\ \partial_y u \end{Bmatrix}

+ fu + g = 0

$$

Let $$\bar{A}$$ be the following coefficient matrix:

$$ \bar{A} =

\begin{bmatrix} a & b\\ b & c \end{bmatrix} $$

Recall (Eqn. 1, pg. 12-3):

$$ \begin{Bmatrix} \partial_x \\ \partial_y \end{Bmatrix}

\left( u \right)

=

\begin{bmatrix} \frac{\partial \bar{x}}{\partial x} & \frac{\partial \bar{y}}{\partial x}\\ \frac{\partial \bar{x}}{\partial y} & \frac{\partial \bar{y}}{\partial y} \end{bmatrix}

\begin{Bmatrix} \partial_{\bar{x}} \\ \partial_{\bar{y}} \end{Bmatrix}

\left( \bar u \right)

=

{\bar J}^T

\begin{Bmatrix} \partial_{\bar{x}} \\ \partial_{\bar{y}} \end{Bmatrix}

\left( \bar u \right)

$$

The general form in matrix notation of this non-linear transformation using the Jacobian matrix is given as:

$$ \left \lfloor \partial_{\bar{x}} \; \partial_{\bar{y}} \right \rfloor

\bar{J}

\left (

\bar{A} \bar{J}^T

\begin{Bmatrix} \partial_{\bar{x}} \\ \partial_{\bar{y}} \end{Bmatrix}

\right )

\left( \bar{u} \right)

+

\left \lfloor d \; e \right \rfloor

\bar{J}^T

\begin{Bmatrix} \partial_{\bar{x}} \\ \partial_{\bar{y}} \end{Bmatrix}

\left( \bar{u} \right)

+ f\left( \bar{u} \right)

+ g = 0

$$

note: $$\bar{A}, d, e, f,$$ and $$g$$ are all functions of $$\bar{x}$$ and $$\bar{y}$$

Expressions of Transformations of Coordinates
--Egm6322.s09.lapetina 17:48, 24 April 2009 (UTC)

Recall the Jacobian Matrix for $$n$$ variables:

$$\bar x_{i}=\bar x_{i}(x_1,...,x_n)$$

$$\bar J=\begin{bmatrix} \partial {\bar x_i}     \\ \partial {\bar x_j} \end{bmatrix} $$.

In two dimensions this can be expressed as:

$$\bar J=\begin{bmatrix} J_{11} & J_{12} \\ J_{21} & J_{22} \end{bmatrix}=\begin{bmatrix} \phi_{x} & \phi_{y} \\ \psi_{x} & \psi_{y} \end{bmatrix}$$

Or, as Lapidus and Pinder show:

$$u_{xx}=u_{\bar x \bar x}(J_{11})^2+2 u_{\bar x \bar y} J_{11} J_{21}+u_{\bar y \bar y}^2$$

To derive Pinder's equations in matrix operator form, we use equation(1) from page 12-3,

$$\begin{Bmatrix} \partial_x \\ \partial_y \end{Bmatrix} \left \lfloor \partial_x \; \partial_y \right \rfloor =\begin{bmatrix} \partial_{xx}     & \partial_{xy}      \\ \partial_{yx}     & \partial_{yy} \end{bmatrix}=\bar J^T \begin{Bmatrix} \partial_{\bar x} \\ \partial_{\bar y} \end{Bmatrix} \left \lfloor \partial_x \; \partial_y \right \rfloor \bar J $$

Note:

$$\left \lfloor \partial_x \; \partial_y \right \rfloor= \left \lfloor \partial_{\bar x} \; \partial_{\bar y} \right \rfloor \bar J$$

$$(\bar A \bar B)^T={\bar B}^T{\bar A}^T$$

$$\alpha={\bar J}^T \begin{bmatrix} \partial_{\bar x \bar x}     & \partial_{\bar x \bar y}      \\ \partial_{\bar y \bar x}     & \partial_{\bar y \bar y}     \end{bmatrix} \bar J $$ To find $$\alpha$$, we must complete some matrix multiplication

$$ \alpha = \begin{bmatrix} J_{11}     & J_{21}      \\ J_{12}     & J_{22} \end{bmatrix} \begin{bmatrix} \partial_{\bar x \bar x}     & \partial_{\bar x \bar y}      \\ \partial_{\bar y \bar x}     & \partial_{\bar y \bar y}  \end{bmatrix} \begin{bmatrix} J_{11}     & J_{12}      \\ J_{21}     & J_{22} \end{bmatrix} = \begin{bmatrix} J_{11}     & J_{21}      \\ J_{12}     & J_{22} \end{bmatrix}

\begin{bmatrix} \partial_{\bar x \bar x} J_{11} +\partial_{\bar x \bar y} J_{21} & \partial_{\bar x \bar x} J_{12} +\partial_{\bar x \bar y} J_{22}    \\ \partial_{\bar y \bar x} J_{11} +\partial_{\bar y \bar y} J_{21}  & \partial_{\bar y \bar x} J_{12} +\partial_{\bar y \bar y} J_{22} \end{bmatrix}

$$

$$ \alpha=

\begin{bmatrix}

\partial_{\bar x \bar x} (J_{11})^2 +2 \partial_{\bar x \bar y} J_{21} J_{11} + \partial_{\bar y \bar y } {J_{21}}^2 & \partial_{\bar x \bar x} J_{11} J_{21} + \partial_{\bar x \bar y} J_{22} J_{11} + \partial_{\bar x \bar y} J_{12} J_{21} + \partial_{\bar y \bar y} J_{22} J_{21} \\

\partial_{\bar x \bar x} J_{11} J_{21} + \partial_{\bar x \bar y} J_{21} J_{12} + \partial_{\bar x \bar y} J_{11} J_{22} + \partial_{\bar y \bar y} J_{21} J_{22} & \partial_{\bar x \bar x} (J_{12})^2 +2 \partial_{\bar x \bar y} J_{12} J_{22} + \partial_{\bar y \bar y } {J_{22}}^2

\end{bmatrix}

$$

$${\alpha}_{11}=(\partial_{\bar x \bar x}(J_{11})^2)+\partial_{\bar x \bar y}J_{11}J_{21}+\partial_{\bar y \bar x}J_{11}J_{21}+\partial_{\bar y \bar x}J_{11}J_{21}+\partial_{\bar y \bar y}(J_{21})^2 $$

Egm6322.s09.three.liu 16:46, 24 April 2009 (UTC)

Assume $$\partial_{\bar x \bar y}=\partial_{\bar y \bar x}$$

With conversions, it can be shown that our expression is equivalent to that of Lapidus and Pinder: $$ \partial_{xx}=\partial_{\bar x \bar x}(J_{11})^2+\partial_{\bar x \bar y}J_{11}J_{21}+\partial_{\bar y \bar y}(J_{21})^2=\partial_{\xi \xi}(\phi_{x})^2+2\partial_{\xi \eta}\phi_{x}\psi_{x}+\partial_{\eta \eta}(\psi_{x})^2$$

This proof completed in one of the homework assignments, found here.

 It is best to report the assignment right here, instead of elsewhere in the report, since that would create some confusion while reading the report. At least you should create in internal link to help the readers jump to the part of the report where the proof was done; see further below. To reedit and resubmit. Egm6322.s09 13:04, 23 February 2009 (UTC)

Link added as requested. --Egm6322.s09.lapetina 20:23, 6 March 2009 (UTC) Homework: Transform the Laplace Equation into polar coordinates, and compare to published results

Egm6322.s09.Three.nav 13:29, 24 April 2009 (UTC)

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

It has the following mathematical form:

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

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

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

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

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

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

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

Differentiating by parts

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

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

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

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

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

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

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

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

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

Evaluating the term in angled brackets for equation(2)

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

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

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

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

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

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

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

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

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

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

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

The Jacobian for Linear and Affine Maps
Egm6322.s09.Three.ge 17:32, 24 April 2009 (UTC)

The Jacobian matrix is a function of $$x$$ and $$y$$ in general.

$$\underline J \left ( x,y \right )$$ in general is not a constant matrix.

However, $$\underline J $$ is a constant matrix for linear and affine maps. In summary,

$$\begin{Bmatrix} \underline{x}\\ \underline{y} \end{Bmatrix} = \underline{E} \begin{Bmatrix} x\\ y \end{Bmatrix} + \underline{F}$$

Where:

$$\underline{J}=\underline{E}$$

An Example of Affine Mapping
Egm6322.s09.bit.sahin 16:23, 24 April 2009 (UTC); Egm6322.s09.Three.ge 17:32, 24 April 2009 (UTC)

Example of affine map.

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

To get the transformation matrix, note that:

$$\overrightarrow{OP}=\overrightarrow{O\overline{O}}+\overrightarrow{\overline{O}\mathbf{P}}$$

Also note:

$$\overrightarrow{O\mathbf{P}}=x{\mathbf{i}}+y{\mathbf{j}}$$

$$\overrightarrow{O\overline{O}}=x_{o}\mathbf{i}+y_{o}\mathbf{j}$$

$$\overrightarrow{O\mathbf{P}}=\bar{x}{\mathbf{i}}+\bar{y}{\mathbf{j}}=\bar{x}(m\mathbf{i}+n\mathbf{j})+\bar{y}(p\mathbf{i}+q\mathbf{j})$$

$$=(m\bar{x}+p\bar{y})\mathbf{i}+(n\bar{x}+q\bar{y})\mathbf{j}$$

where:

$$x=m\bar{x}+p\bar{y}+x_0$$

$$y=n\bar{x}+q\bar{y}+x_0$$

Using the matrix form:

$$\begin{Bmatrix}x\\y\end{Bmatrix}= \begin{bmatrix}m & p\\n & q\end{bmatrix}\begin{Bmatrix}\bar{x}\\\bar{y}\end{Bmatrix}+ \begin{Bmatrix}x_0\\y_0\end{Bmatrix}$$

then,

$$x=x_{0}+\overline{x}m+\overline{y}p$$

$$y=y_{0}+\overline{x}n+\overline{y}q$$

from the equation system $$\overline{x}$$ and $$\overline{y}$$ are obtained as

$$\overline{x}=\frac{q}{\left (mq-np \right )}x-\frac{p}{\left (mq-np \right )}y-\frac{q}{\left (mq-np \right )}x_{0}+\frac{p}{\left (mq-np \right )}y_{0}$$

$$\overline{y}=\frac{m}{\left (mq-np \right )}y-\frac{n}{\left (mq-np \right )}x+\frac{n}{\left (mq-np \right )}x_{0}-\frac{m}{\left (mq-np \right )}y_{0}$$

Recall:

$$\begin{Bmatrix} \overline{x}\\ \overline{y} \end{Bmatrix} = \underline{E} \begin{Bmatrix} x\\ y \end{Bmatrix} + \underline{F}$$

$$\begin{Bmatrix} \overline{x}\\ \overline{y}

\end{Bmatrix} =\begin{bmatrix} e_{11} &e_{12} \\ e_{21} &e_{22} \end{bmatrix}\begin{Bmatrix} x\\ y

\end{Bmatrix}+\begin{Bmatrix} f_{11}\\ f_{21}

\end{Bmatrix}$$

therefore,

$$\overline{x}=e_{11}x+e_{12}y+f_{11}$$

$$\overline{y}=e_{21}x+e_{22}y+f_{21}$$

Hence the coefficient are,

$$e_{11}=\frac{q}{\left (mq-np \right )}$$

$$e_{12}=\frac{-p}{\left (mq-np \right )}$$

$$f_{11}=\frac{py_{0}-qx_{0}}{\left (mq-np \right )}$$

$$e_{21}=\frac{m}{\left (mq-np \right )}$$

$$e_{22}=\frac{-n}{\left (mq-np \right )}$$

$$f_{21}=\frac{nx_{0}-my_{0}}{\left (mq-np \right )}$$

Eventually we obtain $$\underline{E}$$ and $$\underline{F}$$ as following

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

$$\underline{F}=\begin{Bmatrix} \frac{py_{0}-qx_{0}}{\left (mq-np \right )}\\ \frac{nx_{0}-my_{0}}{\left (mq-np \right )} \end{Bmatrix}$$

Exercise 1 below shows this with values for $$n,m,p,q x_0$$ and $$y_0$$.

 Not clear. To redo and resubmit. Re-explain and redo in class if still have problems. Egm6936.s09 02:20, 21 February 2009 (UTC)

Changes needed were made. Egm6322.s09.bit.sahin 15:52, 10 April 2009 (UTC)

Homework Problems
Homework: Find the Jacobian Matrix for the transformation from Cartesian to Polar Coordinate

One example of nonlinear coordinate transformation is from Cartesian Coordinates to Polar coordinates:

e.g.

Cartesian coordinates can be expressed as:$$\left ( x,y \right )$$

while polar coordinates are expressed as:$$(r,\theta)$$.

The relationship between the two coordinates is:

$$x=r\cos\theta=\bar{x}\cos\bar{y}$$

$$y=r\sin\theta=\bar{x}\sin\bar{y}$$

To find J: $$ J=\begin{bmatrix} \frac{\partial {\bar x_i}}   {\partial {x_j}} \end{bmatrix} $$ = $$\begin{bmatrix} J_{11}     & J_{12}     \\ J_{21}     & J_{22} \end{bmatrix}$$

$$x=rcos\Theta =\bar{x_1}cos\bar{x_2}=x_1(\bar{x_1}, \bar{x_2})$$

$$y=rsin\Theta =\bar{x_1}sin\bar{x_2}=x_2(\bar{x_1}, \bar{x_2})$$

$$\bar{x_1}=r=(x^2+y^2)^{1/2}=\bar{x_1}(x_1, x_2)$$

$$\bar{x_2}=\Theta =tan^{-1}(\frac{y}{x})=\bar{x_2}(x_1, x_2)$$

$$J_{11}=\frac{\partial \bar{x}_1}{\partial x}=\frac{\partial r}{\partial x}=(x^2+y^2)^{-1/2}x$$

$$J_{12}=\frac{\partial \bar{x}_2}{\partial r}=\frac{\partial r}{\partial x}=(x^2+y^2)^{-1/2}y$$

$$J_{21}=\frac{\partial \bar{x}_2}{\partial x_1}=\frac{\partial \Theta}{\partial x}=\frac{\partial \tan^{-1}(s)}{\partial s}\frac{\partial s}{\partial x}$$

where: $$s:=\frac{y}{x}$$

It becomes clear that one must take the derivative of $$tan^{-1}(\frac {y} {x})$$.

Recall: $$y=tan^{-1}x\Rightarrow x=tan(y)=\frac{sin(y)}{cos(y)}$$

Find $$\frac{d y}{d x}$$

$$\frac{d y}{d x}=\frac{1}{\frac{d x}{d y}}$$

$$\frac{d x}{d y}=\frac{(cosy)^2-siny(-siny)}{cos^2y}=\frac{1}{cos^2y}=1+tan^2y=1+x^2$$

Therefore,

$$\frac{d y}{d x}=cos^2y$$

$$tan(y)=\frac{sin(y)}{cos(y)}$$

Expressing J in terms of $$\begin{pmatrix}r,&\Theta \end{pmatrix} $$

$$J_{11}=\frac{x} {r}=\frac{rcos\Theta }{r}=cos\Theta$$

$$J_{12}=\frac{y}{\left (x^{2}+y^{2} \right )^{1/2}}=\frac{rsin\theta }{r}=sin\theta $$

$$J_{21}=\frac{1}{1+\left (\frac{y}{x} \right )^{2}}\left (-\frac{y}{x^{2}} \right )=\frac{-y}{x^{2}+y^{2}}=\frac{-rsin\theta }{r^{2}}=-\frac{sin\theta }{r}$$

$$J_{22}=\frac{\partial \overline{x}_{2}}{\partial x_{2}}=\frac{\partial \theta }{\partial y}$$

Remembering that $$\theta =tan^{-1}\left (\frac{y}{x} \right )$$, If we define $$s=y/x$$ then

$$\frac{\partial \theta }{\partial y}=\frac{\partial }{\partial s}tan^{-1}(s)\frac{\partial s}{\partial y}=\frac{1}{1+\left (y/x \right )^{2}}\left (\frac{1}{x} \right )=\frac{x}{x^{2}+y^{2}}=\frac{rcos\theta }{r^{2}}=\frac{cos\theta }{r}$$

As a result, we obtain Jacobian matrix as

$$J= \begin{bmatrix} J_{11}     & J_{12}     \\ J_{21}     & J_{22} \end{bmatrix} = \begin{bmatrix} cos\theta & sin\theta \\ -\frac{sin\theta }{r}&\frac{cos\theta }{r} \end{bmatrix}=\frac{1}{r}\begin{bmatrix} rcos\theta &rsin\theta  \\ -sin\theta &cos\theta \end{bmatrix}$$

 There is an error in the derivative of $$\displaystyle y = \tan^{-1} x$$ above. Egm6322.s09 17:11, 23 February 2009 (UTC)

 Put your reply inside the above comment box, and sign with 4-tilde command. There is still one more "misprint"; see further below for a hint about the same misprint. Egm6322.s09 10:04, 2 April 2009 (UTC)

 This part has been changed. Egm6322.s09.three.liu 14:33, 10 April 2009 (UTC)

Homework: Find Complete Expression for Eqn. 1.2.6, 1.2.7, & 1.2.8 in Lapidus & Pinder [1982] --Egm6322.s09.xyz 16:19, 4 April 2009 (UTC)

Compare the expansion of Equation 1 on page 12-3 to LP (Lapidus & Pinder, 1982), found on page 14 of their book. Translate their notation into ours.

Obtain the first partial derivative under transform of coordinates: Eq(1) in P12-3. (using chain rule)

In LP, P14: Obtain second partial derivative (using chain rule), translate it into new coordinate (ours).

Where ours: $$(\bar x_1, \bar x_2)=(\bar x, \bar y)$$

$$\bar {x_i}=\bar{x_i} (x_1,x_2) $$    (i=1,2,...)

But for Lapidus and Pinder: $$\xi=\phi(x,y)$$

$$\eta=\psi(x,y)$$

LP (1.2.5 ab): First partial derivative

LP (1.2.6-8): Second partial derivative

Obtain the complete form:

$$u_{xx}=u_{\xi\xi}(\phi_{x})^2+2 u_{\xi \eta} \phi_{x} \psi_{x}+u_{\eta \eta}+(\psi_{x})^2+...$$

Recall that a 2nd order PDE with two (2) independent variables has the following form (see L&P, eqn. 1.2.1):

$$a \left( \cdot \right) u_{xx} + 2b \left( \cdot \right) u_{xy} + c \left( \cdot \right) u_{yy} + d \left( \cdot \right) u_x + e \left( \cdot \right) u_y + f \left( \cdot \right) u + g \left( \cdot \right) = 0$$

where $$\xi = \phi \left(x,y \right)$$ and $$\eta = \psi \left(x,y \right)$$

Chain rule is used to develop the following expressions for the first and second order derivatives with respect to x,y:

$$\blacktriangleright u_x = u_{\xi}\phi_x + u_{\eta}\psi_x$$

$$\blacktriangleright u_y = u_{\xi}\phi_y + u_{\eta}\psi_y$$

$$\blacktriangleright u_{xx} = u_{\xi\xi}\phi_x^2 + 2u_{\xi\eta}\phi_x\psi_x + u_{\eta\eta}\psi_x^2 + u_{\xi}\phi_{xx} + u_{\eta}\psi_{xx}$$ (Eqn. 1.2.6)

$$\blacktriangleright u_{xy} = u_{\xi\xi}\phi_x\phi_y + u_{\xi\eta}\left( \phi_x\psi_y + \phi_y\psi_x \right) + u_{\eta\eta}\psi_x\psi_y + u_{\xi}\phi_{xy} + u_{\eta}\psi_{xy}$$ (Eqn. 1.2.7)

$$\blacktriangleright u_{yy} = u_{\xi\xi}\phi_y^2 + 2u_{\xi\eta}\phi_y\psi_y + u_{\eta\eta}\psi_y^2 + u_{\xi}\phi_{yy} + u_{\eta}\psi_{yy}$$ (Eqn. 1.2.8)

Solution:

The complete form of the equations 1.2.6 to 1.2.8 from Lapidus and Pinder:

Let,

$$  \xi =\phi (x,y)$$

and

$$ \eta =\psi (x,y) $$

$$\therefore $$ by chain rule we have

$$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial x} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial x}$$

and

$$\frac{\partial u}{\partial y}=\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial y}$$

EXPRESSION FOR $$\frac{\partial^2 u}{\partial x^2}$$

We Know that

$$\frac{\partial^2 u}{\partial x^2}=\frac{\partial }{\partial x}\left (\frac{\partial u}{\partial x} \right )$$ $$ \therefore$$ from chain rule we have

fex $$ \frac{\partial^2 u}{\partial x^2}=\frac{\partial }{\partial \xi}\left (\frac{\partial u}{\partial x} \right )\frac{\partial \phi}{\partial x} + \frac{\partial }{\partial \eta}\left (\frac{\partial u}{\partial x} \right )\frac{\partial \psi}{\partial x}$$

$$\frac{\partial^2 u}{\partial x^2}=\frac{\partial }{\partial \xi}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial x} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial x} \right )\frac{\partial \phi}{\partial x} + \frac{\partial }{\partial \eta}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial x} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial x} \right )\frac{\partial \psi}{\partial x} $$

Let,

$$\frac{\partial }{\partial \xi}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial x} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial x} \right )\frac{\partial \phi}{\partial x} =I$$

and

$$\frac{\partial }{\partial \eta}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial x} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial x} \right )\frac{\partial \psi}{\partial x}=II

$$

Differentiating $$I$$ We have

$$I=\frac{\partial }{\partial \xi}\left(\frac{\partial u}{\partial \xi}\right)\left(\frac{\partial \phi}{\partial x}\right)^2 + \frac{\partial }{\partial \xi}\left(\frac{\partial \phi}{\partial x}\right )\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial x}$$

$$ + \frac{\partial }{\partial \xi}\left(\frac{\partial u}{\partial \eta}\right)\frac{\partial \psi}{\partial x}\frac{\partial \phi}{\partial x} + \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \xi}\left(\frac{\partial \psi}{\partial x}\right)\frac{\partial \phi}{\partial x} $$

and

$$II=\frac{\partial }{\partial \eta}\left(\frac{\partial u}{\partial \xi}\right)\frac{\partial \phi}{\partial x}\frac{\partial \psi}{\partial x} + \frac{\partial }{\partial \eta}\left(\frac{\partial \phi}{\partial x}\right )\frac{\partial u}{\partial \xi}\frac{\partial \psi}{\partial x} $$

$$+\frac{\partial }{\partial \eta}\left(\frac{\partial u}{\partial \eta}\right)\left(\frac{\partial \psi}{\partial x}\right)^2 + \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \eta}\left(\frac{\partial \psi}{\partial x}\right)\frac{\partial \psi}{\partial x}

$$

$$\Rightarrow I=\frac{\partial^2 u}{\partial \xi^2}\left(\frac{\partial \phi}{\partial x}\right)^2 + \frac{\partial }{\partial \xi}\left(\frac{\partial \phi}{\partial x}\right )\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial x} + \frac{\partial^2 u}{\partial \xi{\partial \eta}}\frac{\partial \psi}{\partial x}\frac{\partial \phi}{\partial x} + \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \xi}\left(\frac{\partial \psi}{\partial x}\right)\frac{\partial \phi}{\partial x}

$$

$$\Rightarrow II=\frac{\partial^2 u}{\partial \xi{\partial \eta}}\frac{\partial \psi}{\partial x}\frac{\partial \phi}{\partial x} + \frac{\partial }{\partial \eta}\left(\frac{\partial \phi}{\partial x}\right )\frac{\partial u}{\partial \xi}\frac{\partial \psi}{\partial x} + \frac{\partial^2 u}{\partial \eta^2}\left(\frac{\partial \psi}{\partial x}\right)^2 + \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \eta}\left(\frac{\partial \psi}{\partial x}\right)\frac{\partial \psi}{\partial x} $$

The complete expansion of $$u_{xx}=$$

$$ I+II=\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2  u}{\partial \xi^2}\left(\frac{\partial \phi}{\partial x}\right)^2+2\frac{\partial^2 u}{\partial \xi{\partial \eta}}\frac{\partial \psi}{\partial x}\frac{\partial \phi}{\partial x}+\frac{\partial^2 u}{\partial \eta^2}\left(\frac{\partial \psi}{\partial x}\right)^2$$

$$+\frac{\partial }{\partial \xi}\left(\frac{\partial \phi}{\partial x}\right )\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial x} +\frac{\partial u}{\partial \eta}\frac{\partial }{\partial \xi}\left(\frac{\partial \psi}{\partial x}\right)\frac{\partial \phi}{\partial x}+\frac{\partial }{\partial \eta}\left(\frac{\partial \phi}{\partial x}\right )\frac{\partial u}{\partial \xi}\frac{\partial \psi}{\partial x}+\frac{\partial u}{\partial \eta}\frac{\partial }{\partial \eta}\left(\frac{\partial \psi}{\partial x}\right)\frac{\partial \psi}{\partial x}$$

EXPRESSION FOR $$\frac{\partial^2 u}{\partial y^2}$$

Similarly,

$$\frac{\partial^2 u}{\partial y^2}=\frac{\partial }{\partial y}\left (\frac{\partial u}{\partial y} \right )$$ $$ \therefore$$ from chain rule we have

$$ \frac{\partial^2 u}{\partial y^2}=\frac{\partial }{\partial \xi}\left (\frac{\partial u}{\partial y} \right )\frac{\partial \phi}{\partial y} + \frac{\partial }{\partial \eta}\left (\frac{\partial u}{\partial y} \right )\frac{\partial \psi}{\partial y}$$

$$\frac{\partial^2 u}{\partial y^2}=\frac{\partial }{\partial \xi}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial y} \right )\frac{\partial \phi}{\partial y} + \frac{\partial }{\partial \eta}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial y} \right )\frac{\partial \psi}{\partial y} $$

Let,

$$\frac{\partial }{\partial \xi}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial y} \right )\frac{\partial \phi}{\partial y} =I$$

and

$$\frac{\partial }{\partial \eta}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial y} \right )\frac{\partial \psi}{\partial y}=II

$$

Differentiating $$I$$ We have

$$I=\frac{\partial }{\partial \xi}\left(\frac{\partial u}{\partial \xi}\right)\left(\frac{\partial \phi}{\partial y}\right)^2 + \frac{\partial }{\partial \xi}\left(\frac{\partial \phi}{\partial y}\right )\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y}$$

$$ + \frac{\partial }{\partial \xi}\left(\frac{\partial u}{\partial \eta}\right)\frac{\partial \psi}{\partial y}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \xi}\left(\frac{\partial \psi}{\partial y}\right)\frac{\partial \phi}{\partial y} $$

and

$$II=\frac{\partial }{\partial \eta}\left(\frac{\partial u}{\partial \xi}\right)\frac{\partial \phi}{\partial y}\frac{\partial \psi}{\partial y} + \frac{\partial }{\partial \eta}\left(\frac{\partial \phi}{\partial y}\right )\frac{\partial u}{\partial \xi}\frac{\partial \psi}{\partial y} $$

$$+\frac{\partial }{\partial \eta}\left(\frac{\partial u}{\partial \eta}\right)\left(\frac{\partial \psi}{\partial y}\right)^2 + \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \eta}\left(\frac{\partial \psi}{\partial y}\right)\frac{\partial \psi}{\partial y}

$$

$$\Rightarrow I=\frac{\partial^2 u}{\partial \xi^2}\left(\frac{\partial \phi}{\partial y}\right)^2 + \frac{\partial }{\partial \xi}\left(\frac{\partial \phi}{\partial y}\right )\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y} + \frac{\partial^2 u}{\partial \xi{\partial \eta}}\frac{\partial \psi}{\partial y}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \xi}\left(\frac{\partial \psi}{\partial y}\right)\frac{\partial \phi}{\partial y}

$$

$$\Rightarrow II=\frac{\partial^2 u}{\partial \xi{\partial \eta}}\frac{\partial \psi}{\partial y}\frac{\partial \phi}{\partial y} + \frac{\partial }{\partial \eta}\left(\frac{\partial \phi}{\partial y}\right )\frac{\partial u}{\partial \xi}\frac{\partial \psi}{\partial y} + \frac{\partial^2 u}{\partial \eta^2}\left(\frac{\partial \psi}{\partial y}\right)^2 + \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \eta}\left(\frac{\partial \psi}{\partial y}\right)\frac{\partial \psi}{\partial y} $$

The complete eypansion of $$u_{yy}=$$

$$ I+II=\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2  u}{\partial \xi^2}\left(\frac{\partial \phi}{\partial y}\right)^2+2\frac{\partial^2 u}{\partial \xi{\partial \eta}}\frac{\partial \psi}{\partial y}\frac{\partial \phi}{\partial y}+\frac{\partial^2 u}{\partial \eta^2}\left(\frac{\partial \psi}{\partial y}\right)^2$$

$$+\frac{\partial }{\partial \xi}\left(\frac{\partial \phi}{\partial y}\right )\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y} +\frac{\partial u}{\partial \eta}\frac{\partial }{\partial \xi}\left(\frac{\partial \psi}{\partial y}\right)\frac{\partial \phi}{\partial y}+\frac{\partial }{\partial \eta}\left(\frac{\partial \phi}{\partial y}\right )\frac{\partial u}{\partial \xi}\frac{\partial \psi}{\partial y}+\frac{\partial u}{\partial \eta}\frac{\partial }{\partial \eta}\left(\frac{\partial \psi}{\partial y}\right)\frac{\partial \psi}{\partial y}$$

EXPRESSION FOR $$\frac{\partial^2 u}{\partial x\partial y}$$

$$\frac{\partial^2 u}{\partial x\partial y}=\frac{\partial }{\partial x}\left (\frac{\partial u}{\partial y} \right )$$

$$\Rightarrow \frac{\partial^2 u}{\partial x\partial y}=\frac{\partial }{\partial \xi}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial y} \right )\frac{\partial \phi}{\partial x} + \frac{\partial }{\partial \eta}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial y} \right )\frac{\partial \psi}{\partial x}$$

$$I=\frac{\partial }{\partial \xi}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial y} \right )\frac{\partial \phi}{\partial x}$$

$$II=\frac{\partial }{\partial \eta}\left (\frac{\partial u}{\partial \xi}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \eta}\frac{\partial \psi}{\partial y} \right )\frac{\partial \psi}{\partial x}$$

$$\Rightarrow I=\frac{\partial^2 u }{\partial \xi^2}\frac{\partial \phi}{\partial x}\frac{\partial \phi}{\partial y} + \frac{\partial u}{\partial \xi}\frac{\partial }{\partial \xi}\left (\frac{\partial \phi}{\partial y} \right )\frac{\partial \phi}{\partial x} + \frac{\partial^2 u}{\partial \xi\partial \eta}\frac{\partial \psi}{\partial x}\frac{\partial \phi}{\partial x} + \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \xi}\left (\frac{\partial \psi}{\partial y} \right )\frac{\partial \phi}{\partial x} $$

$$II=\frac{\partial^2 u}{\partial \xi\partial \eta}\frac{\partial \psi}{\partial x}\frac{\partial \phi}{\partial x} + \frac{\partial u}{\partial \xi}\frac{\partial }{\partial \eta}\left (\frac{\partial \phi}{\partial y} \right )\frac{\partial \psi}{\partial x} + \frac{\partial^2 u }{\partial \eta^2}\frac{\partial \psi}{\partial y}\frac{\partial \psi}{\partial x} + \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \eta}\left (\frac{\partial \psi}{\partial y} \right )\frac{\partial \psi}{\partial x} $$

$$\Rightarrow \frac{\partial^2 u}{\partial x\partial y}=I+II=\frac{\partial^2 u }{\partial \xi^2}\frac{\partial \phi}{\partial x}\frac{\partial \phi}{\partial y}+ \frac{\partial^2 u}{\partial \xi\partial \eta}\left(\frac{\partial \psi}{\partial x}\frac{\partial \phi}{\partial x}+\frac{\partial \psi}{\partial x}\frac{\partial \phi}{\partial x}\right)+\frac{\partial^2 u }{\partial \eta^2}\frac{\partial \psi}{\partial y}\frac{\partial \psi}{\partial x}$$

$$+\frac{\partial u}{\partial \xi}\frac{\partial }{\partial \xi}\left (\frac{\partial \phi}{\partial y} \right )\frac{\partial \phi}{\partial x}+ \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \xi}\left (\frac{\partial \psi}{\partial y} \right )\frac{\partial \phi}{\partial x}+ \frac{\partial u}{\partial \xi}\frac{\partial }{\partial \eta}\left (\frac{\partial \phi}{\partial y} \right )\frac{\partial \psi}{\partial x}+ \frac{\partial u}{\partial \eta}\frac{\partial }{\partial \eta}\left (\frac{\partial \psi}{\partial y} \right )\frac{\partial \psi}{\partial x}$$

The matrix form of the above expressions

We Have

$$\begin{bmatrix} \frac{\partial }{\partial x} \\ \frac{\partial }{\partial y} \end{bmatrix} =J^T\begin{bmatrix} \frac{\partial }{\partial \overline x} \\ \frac{\partial }{\partial \overline y} \end{bmatrix}$$

Let this be 1

and

$$\begin{bmatrix} \frac{\partial }{\partial x} \\ \frac{\partial }{\partial y} \end{bmatrix}

\begin{bmatrix} \frac{\partial }{\partial x} & \frac{\partial }{\partial y} \end{bmatrix}= \begin{bmatrix} \partial_{xx} &\partial_{xy} \\ \partial_{yx} &\partial_{yy} \end{bmatrix}$$

Let this be 2

Also

$$\because \left (AB \right )^T=B^TA^T$$

$$\begin{bmatrix} \partial_x & \partial_y \end{bmatrix}= \begin{bmatrix} \partial_ \bar x & \partial_\bar y \end{bmatrix}J$$

Let this be 3

Applying equation 2 and 3 in equation 1 ,we have

$$\begin{bmatrix} \partial_x \\ \partial_y \end{bmatrix} \begin{bmatrix} \partial_x & \partial_y \end{bmatrix} = J^T \begin{bmatrix} \partial_\bar x \\ \partial_\bar y \end{bmatrix} \begin{bmatrix} \partial_\bar x & \partial_\bar y \end{bmatrix} J$$

$$\Rightarrow \begin{bmatrix} \partial_{xx} &\partial_{xy} \\ \partial_{yx} &\partial_{yy} \end{bmatrix} =\begin{bmatrix} J_{11} & J_{21}\\ J_{12} & J_{22} \end{bmatrix} \begin{bmatrix} \partial_{\bar x\bar x} &\partial_{\bar x\bar y} \\ \partial_{\bar y\bar x} &\partial_{\bar y\bar y} \end{bmatrix} \begin{bmatrix} J_{11} &J_{12} \\ J_{11}& J_{22} \end{bmatrix}$$

$$\Rightarrow \begin{bmatrix} \partial_{xx} &\partial_{xy} \\ \partial_{yx} &\partial_{yy} \end{bmatrix}$$

$$= \begin{bmatrix}

\partial_{\bar x \bar x} (J_{11})^2 +2 \partial_{\bar x \bar y} J_{21} J_{11} + \partial_{\bar y \bar y } {J_{21}}^2 & \partial_{\bar x \bar x} J_{11} J_{21} + \partial_{\bar x \bar y} J_{22} J_{11} + \partial_{\bar x \bar y} J_{12} J_{21} + \partial_{\bar y \bar y} J_{22} J_{21} \\

\partial_{\bar x \bar x} J_{11} J_{21} + \partial_{\bar x \bar y} J_{21} J_{12} + \partial_{\bar x \bar y} J_{11} J_{22} + \partial_{\bar y \bar y} J_{21} J_{22} & \partial_{\bar x \bar x} (J_{12})^2 +2 \partial_{\bar x \bar y} J_{12} J_{22} + \partial_{\bar y \bar y } {J_{22}}^2

\end{bmatrix}$$

$$\therefore u_{xx}=\partial_{\bar x \bar x}(J_{11})^2+\partial_{\bar x \bar y}J_{11}J_{21}+\partial_{\bar y \bar y}(J_{21})^2=\partial_{\xi \xi}(\phi_{x})^2+2\partial_{\xi \eta}\phi_{x}\psi_{x}+\partial_{\eta \eta}(\psi_{x})^2$$

$$u_{xy}=\partial_{\bar x \bar x} J_{11} J_{21} + \partial_{\bar x \bar y} J_{22} J_{11} + \partial_{\bar x \bar y} J_{12} J_{21} + \partial_{\bar y \bar y} J_{22} J_{21} =u_{\xi\xi}\phi_x\phi_y + u_{\xi\eta}\left( \phi_x\psi_y + \phi_y\psi_x \right) + u_{\eta\eta}\psi_x\psi_y $$

$$u_{yy}=\partial_{\bar x \bar x} (J_{12})^2 +2 \partial_{\bar x \bar y} J_{12} J_{22} + \partial_{\bar y \bar y } {J_{22}}^2 =u_{\xi\xi}\phi_y^2 + 2u_{\xi\eta}\phi_y\psi_y + u_{\eta\eta}\psi_y^2 + $$

Egm6322.s09.bit.gk 20:32, 24 April 2009 (UTC)

 The proof by the matrix-operator method was incomplete (wrong); review the lecture notes. Egm6322.s09 13:04, 23 February 2009 (UTC)

 As mentioned in class, there are missing first-order terms in the proof by the matrix-operator method as compared to using chain-rule done further above. Needs correction. Egm6322.s09 16:30, 2 April 2009 (UTC)

Pierre-Simon Laplace (1749-1827)

Egm6322.s09.Three.nav 13:32, 24 April 2009 (UTC)

Pierre-Simon Laplace was born on 23 March, 1747 in Beaumont-en-Auge, Normandy, France. At the age of 16, while in university studying theology, he discovered a natural talent and a deep love for mathematics. He then set sail for Paris to become a student under the guidance of D'alembert. At the tender age of 19, he began produced a remarkable stream of Mathematical papers. His research was as diverse as it was trendsetting. In 1773, at the age of 24, he was elected an adjoint in the Académie des Sciences (French Academy of Sciences). It was here that he maintained many superior posts and was widely considered one of the Academy's best mathematicians.

The Laplace Equation is first mentioned in Laplace's most important work, Traité de Mécanique Céleste. Though we now name this equation after Laplace, it was actually known well before the time of Laplace. Laplace died on the morning of March 5, 1827. The Academy, seldom known for cancelling any of its meetings, did so on that day as a mark of respect for him.

Laplace

Homework: Transform the Laplace Equation into polar coordinates, and compare to published results

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

It has the following mathematical form:

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

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

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

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

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

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

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

Differentiating by parts

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

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

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

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

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

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

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

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

$$\Rightarrow\ (2)= \left \lfloor \partial_{x} \ \partial_{y} \right \rfloor J^{T} \begin{Bmatrix} \partial_{r}\\ \partial_{\theta} \end{Bmatrix} \left(u\right)$$

Evaluating the term in angled brackets for equation(2)

$$\left \lfloor \partial_{x} \ \partial_{y} \right \rfloor J^{T} \begin{Bmatrix} \partial_{r}\\ \partial_{\theta} \end{Bmatrix} \left(u\right)= \begin{bmatrix} \left(\partial_{x}C+\partial_{y}S \right) \ \left (-\partial_{x}\frac {S}{r}+\partial_{y}\frac{C}{r} \right) \end{bmatrix}$$

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

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

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

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

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

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

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

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

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

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

 If the terms in the angle brackets are kept fixed, then why do you differentiate the following:

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

which should be better written as follows:

\displaystyle \left\lfloor \partial_{x} \ \partial_{y} \right\rfloor \mathbf J ^{T} =  \left\lfloor \left(	 \partial_{x}C+\partial_{y}S     \right) \       \left(         -\partial_{x}\frac {S}{r}+\partial_{y}\frac{C}{r}      \right) \right\rfloor $$ In the expression

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

just omit the angle brackets, and write as follows instead

\displaystyle \Rightarrow \    (2)   =   \left(      \left\lfloor	 \partial_{x} \ \partial_{y}      \right\rfloor      \mathbf J ^{T}   \right) \left\{ \begin{array}{l} \partial_{r} \\	 \partial_{\theta} \end{array} \right\} \left( u \right) $$ Then it makes sense to carry out the differentiation in

\displaystyle \left(     \left\lfloor	 \partial_{x} \ \partial_{y}      \right\rfloor      \mathbf J ^{T}   \right) $$

The necessary corrections have been made. By saying "the terms in the angled brackets are kept fixed", we mean to say that these terms are not subject to differentiation.

Egm6322.s09.Three.nav 19:00, 6 March 2009 (UTC)Egm6322.s09.Three.nav Egm6322.s09 17:51, 23 February 2009 (UTC)

 Move your reply to below my signature, then delete this comment box. Egm6322.s09 00:31, 2 April 2009 (UTC)

Rotation of Coordinates

Egm6322.s09.Three.nav 13:31, 24 April 2009 (UTC)

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

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

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

Then

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

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

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

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

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

Hence the Jacobian ( J ) is given by

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

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

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

 The above problem could be done simply and elegantly by the more general method described above in an example of affine mapping. Egm6322.s09 17:11, 23 February 2009 (UTC) Do not exactly understand how to apply the affine mapping to this example. How do you determine m,n,p,q? Egm6322.s09.Three.nav 19:22, 6 March 2009 (UTC)Egm6322.s09.Three.nav

 Add a figure to better explain the problem. The simpler method had been explained in class on several occasions. Ask me if not clear. Egm6322.s09 10:04, 2 April 2009 (UTC)

Simpler method has been implemented. Egm6322.s09.Three.nav 15:17, 10 April 2009 (UTC)

Example, Rotation and Translation
 The above section title does not make sense in view of the content below. The goal of what is done below is to obtain the expression of the second derivatives. Egm6322.s09 00:31, 2 April 2009 (UTC)

$$\alpha=\begin{Bmatrix} \partial_{x}\\ \partial_{y} \end{Bmatrix} \left \lfloor \partial_{x} \ \partial_{y} \right \rfloor$$

$$\begin{Bmatrix} \partial_{x}\\ \partial_{y} \end{Bmatrix}\left ( \lfloor \partial_{\overline{x}} \; \partial_{\overline{y}} \right \rfloor  \underline{J}(x,y))$$

If J is not a constant, take the product rule of the above function.

$$\alpha = \underline{J}^{T}\begin{Bmatrix} \partial_{\underline{x}}\\ \partial_{\underline{y}} \end{Bmatrix} \left \lfloor \partial_{\overline{x}} \; \partial_{\overline{y}} \right \rfloor \underline{J}+ \begin{Bmatrix} \partial_{x}\\ \partial_{y} \end{Bmatrix} \left \lfloor \partial_{\overline{x}} \; \partial_{\overline{y}} \right \rfloor \underline{J}$$

Let

$$ \ \beta=\underline{J}^{T}\begin{Bmatrix} \partial_{\underline{x}}\\ \partial_{\underline{y}} \end{Bmatrix} \left \lfloor \partial_{\overline{x}} \; \partial_{\overline{y}} \right \rfloor \underline{J}$$ and $$ \gamma= \begin{Bmatrix} \partial_{x}\\ \partial_{y} \end{Bmatrix} \left \lfloor \partial_{\overline{x}} \; \partial_{\overline{y}} \right \rfloor \underline{J}$$

In $$\beta$$ $$ \underline J $$ is treated as a constant while in $$\gamma$$ the operator $$\left \lfloor \partial_{\overline{x}} \;  \partial_{\overline{y}} \right \rfloor$$ is treated as a constant. Thus $$\beta$$ is 2nd order on the unknown function u, and $$\gamma$$ is a 1st order on u. This results in a non-zero $$\gamma$$ matrix of:

$$\gamma= \begin{bmatrix} \partial_{\overline{x}} \phi_{xx}+\partial_{\overline{y}} \psi_{xx} & \partial_{\overline{x}} \phi_{yx}+\partial_{\overline{y}} \psi_{yx}\\ \partial_{\overline{x}} \phi_{xy}+\partial_{\overline{y}} \psi_{xy} & \partial_{\overline{x}} \phi_{yy}+\partial_{\overline{y}} \psi_{yy}\end{bmatrix}$$

If J is a constant then the $$\gamma$$ matrix is equal to zero.

Homework: The Complete Form Of $$\gamma$$ --EGM6322.S09.TIAN 17:37, 24 April 2009 (UTC)

$$\gamma =$$ $$\begin{bmatrix} ( \partial_{\bar{x}}\phi_{xx} + \partial_{\bar{y}})\psi_{xx} & ( \partial_{\bar{x}}\phi_{xy} + \partial_{\bar{y}})\psi_{xy}\\ ( \partial_{\bar{x}}\phi_{xy} + \partial_{\bar{y}})\psi_{xy} & ( \partial_{\bar{x}}\phi_{xy} + \partial_{\bar{y}}\psi_{xy}) \end{bmatrix} $$

 Why was the above collapsible box needed? The expression for $$\displaystyle \gamma$$ is almost the same as the same expression for $$\displaystyle \gamma$$ above the collapsible box. Egm6322.s09 00:31, 2 April 2009 (UTC)

=Exercises =

Exercise 1
--Egm6322.s09.lapetina 01:59, 17 April 2009 (UTC)

Given $${\overline {\mathbf {i}}} = m \mathbf{i}+ n \mathbf{j} =2 \mathbf{i} + 1 \mathbf{j}$$ and $$\overline{\mathbf {j}} = p \mathbf{i}+ q \mathbf{j} =1 \mathbf{i} + 5 \mathbf{j}$$

and that

$$\left ( x_0, y_0 \right )= \left ( 3, 4 \right )$$, map $$\left ( \bar x,\bar y \right ) $$.

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #006600; background-color: #CCFFFF; text-align: left;"> Stick to the notation for tensors (underline = boldface) used in lectures, i.e., $$  \displaystyle \overline \mathbf i  = m  \mathbf i   + n  \mathbf j $$ and $$  \displaystyle \overline \mathbf j  = p  \mathbf i   + q  \mathbf j $$ Egm6322.s09 00:31, 2 April 2009 (UTC)

Edit made a second time.

--Egm6322.s09.lapetina 20:57, 10 April 2009 (UTC)

This answer will take the form:

$$ \begin{Bmatrix} x \\ y \end{Bmatrix}

=

\mathbf E

\begin{Bmatrix} \bar x \\ \bar y \end{Bmatrix}

+

\mathbf F $$

This is an affine map, because 0 does not map to 0. In affine maps, $$\bar E$$ is equivalent to the Jacobian, where

$$ \mathbf J= \begin{bmatrix} \frac {\partial \bar x}{\partial x} & \frac {\partial \bar x}{\partial y}\\ \frac {\partial \bar y}{\partial x} & \frac {\partial \bar y}{\partial y} \end{bmatrix} $$

Therefore:

$$ \mathbf J= \frac{1}{mq-np}

\begin{bmatrix} q & -p\\ -n & m \end{bmatrix}

= \frac {1}{9} \begin{bmatrix} 5 & -1\\ -1 & 2 \end{bmatrix}

$$

and

$$\mathbf F =- \mathbf J

\begin{Bmatrix} x_0 \\ y_0 \end{Bmatrix} $$

Therefore:

$$ \begin{Bmatrix} \bar x \\ \bar y \end{Bmatrix}

= \mathbf J

\left [ \begin{Bmatrix} x \\ y \end{Bmatrix}

-

\begin{Bmatrix} x_0 \\ y_0 \end{Bmatrix}

\right ]

$$

or

$$ \begin{Bmatrix} \bar x \\ \bar y \end{Bmatrix}

=

\begin{bmatrix} \frac{5}{9} & \frac{-1}{9}\\ \frac{-1}{9} & \frac{2}{9} \end{bmatrix}

\left [ \begin{Bmatrix} x \\ y \end{Bmatrix}

-

\begin{Bmatrix} 3 \\ 4 \end{Bmatrix}

\right ]

$$

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Why do you want to use the above form? There is a simple and elegant way to do this problem. Egm6322.s09 17:11, 23 February 2009 (UTC)

More elegant way included. Old form is still included.

--Egm6322.s09.lapetina 14:05, 26 February 2009 (UTC)

Edited once again

--Egm6322.s09.lapetina 04:59, 27 March 2009 (UTC)

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Identify the matrix $$\displaystyle \bar J$$, i.e., give its numerical value. Egm6322.s09 17:11, 23 February 2009 (UTC)

$$J$$ and $${\bar J}^{-1}$$ restated below. --Egm6322.s09.lapetina 14:05, 26 February 2009 (UTC)

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> It is better to keep this exercise directly after the section on an example of affine mapping. above. Egm6936.s09 11:42, 26 February 2009 (UTC)

Link added. --Egm6322.s09.lapetina 20:39, 6 March 2009 (UTC)

Exercise 2
2 Dimensional Laplacian in polar coordinates:

$$\displaystyle {\rm div}( {\rm grad} \, u )=u_{xx}+u_{yy}=:\alpha$$

pag 13-1 Eq.(1) 2nd order linear PDE in the new coordinate $$(\bar{x}, \bar{y})$$ under non-linear coordinate transformation

$$\alpha = \left \lfloor \partial_{\bar{x}} \; \partial_{\bar{y}} \right \rfloor \partial_{\bar{x}}\partial_{\bar{y}}$$ $$\mathbf{J} \mathbf{1}\mathbf{J^T}\begin{Bmatrix}\partial_{\bar{x}}\\\partial_{\bar{y}}\end{Bmatrix}$$

where $$\mathbf{1}$$ is the identity matrix

Find the J pag 14-2

$$x=rcos\theta =\bar{x_1}cos\bar{x_2}=x_1(\bar{x_1}, \bar{x_2})$$

$$y=rsin\theta =\bar{x_1}sin\bar{x_2}=x_2(\bar{x_1}, \bar{x_2})$$

$$\bar{x_1}=r=(x^2+y^2)^{1/2}=\bar{x_1}(x_1, x_2)$$

$$\bar{x_2}=\theta =tan^{-1}(\frac{y}{x})=\bar{x_2}(x_1, x_2)$$

$$J_{11}=\frac{\partial \bar{x}_1}{\partial x}=\frac{\partial r}{\partial x}=(x^2+y^2)^{-1/2}x$$

$$J_{12}=\frac{\partial \bar{x}_2}{\partial r}=\frac{\partial r}{\partial x}=(x^2+y^2)^{-1/2}y$$

$$J_{21}=\frac{\partial \bar{x}_2}{\partial x_1}=\frac{\partial \theta}{\partial x}=\frac{\partial \tan^{-1}(s)}{\partial s}\frac{\partial s}{\partial x}$$

where: $$s:=\frac{y}{x}$$

Recall: $$y=tan^{-1}x\Rightarrow x=tan(y)=\frac{sin(y)}{cos(y)}$$

Find $$\frac{d y}{d x}$$

$$\frac{d y}{d x}=\frac{1}{\frac{d x}{d y}}$$

$$\frac{d x}{d y}=\frac{(cosy)^2-siny(-siny)}{cos^2y}=\frac{1}{cos^2y}=1+tan^2y=1+x^2$$

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Same error as above. Also, keep these related topics together in one place to ease the reading; one of the goals in reediting report R3. Egm6322.s09 17:11, 23 February 2009 (UTC)

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> This part has been changed. Egm6322.s09.three.liu 14:33, 10 April 2009 (UTC)

$$\frac{d y}{d x}=cos^2y$$

$$tan(y)=\frac{sin(y)}{cos(y)}$$

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #006600; background-color: #CCFFFF; text-align: left;"> Same error as above. Egm6322.s09 00:31, 2 April 2009 (UTC)

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #006600; background-color: #CCFFFF; text-align: left;"> This part has been changed. Egm6322.s09.three.liu 14:33, 10 April 2009 (UTC)

Expressing J in terms of $$\begin{pmatrix}r,\theta \end{pmatrix} $$

Egm6322.s09.bit.sahin 16:29, 24 April 2009 (UTC)

$$J_{11}=\frac{x} {r}=\frac{rcos\theta }{r}=cos\theta$$

$$J_{12}=\frac{y}{\left (x^{2}+y^{2} \right )^{1/2}}=\frac{rsin\theta }{r}=sin\theta $$

$$J_{21}=\frac{1}{1+\left (\frac{y}{x} \right )^{2}}\left (-\frac{y}{x^{2}} \right )=\frac{-y}{x^{2}+y^{2}}=\frac{-rsin\theta }{r^{2}}=-\frac{sin\theta }{r}$$

$$J_{22}=\frac{\partial \overline{x}_{2}}{\partial x_{2}}=\frac{\partial \theta }{\partial y}$$

Remembering that $$\theta =tan^{-1}\left (\frac{y}{x} \right )$$, If we define $$s=y/x$$ then

$$\frac{\partial \theta }{\partial y}=\frac{\partial }{\partial s}tan^{-1}(s)\frac{\partial s}{\partial y}=\frac{1}{1+\left (y/x \right )^{2}}\left (\frac{1}{x} \right )=\frac{x}{x^{2}+y^{2}}=\frac{rcos\theta }{r^{2}}=\frac{cos\theta }{r}$$

As a result, we obtain Jacobian matrix as

$$\underline{J}=\begin{bmatrix} cos\theta & sin\theta \\ -\frac{sin\theta }{r}&\frac{cos\theta }{r} \end{bmatrix}=\frac{1}{r}\begin{bmatrix} rcos\theta &rsin\theta  \\ -sin\theta &cos\theta \end{bmatrix}$$

In addition to this method, a simpler method can be used to determine the Jacobian matrix directly in polar coordinates.

$$x_{i}=x_{i}\left (\overline{x}_{1},\overline{x}_{2} \right )=x_{i}\left (r,\theta \right )$$

Instead of inversing the equation, differentiating $$\frac{\partial x_{i}}{\partial \overline{x}_{j}}$$ is simpler (e.g., $$\frac{\partial x}{\partial r},\frac{\partial x}{\partial \theta})$$.

$$\underline{A}=\begin{bmatrix} \frac{\partial x_{i}}{\partial \overline{x}_{j}} \end{bmatrix} =\begin{bmatrix} cos\theta &-rsin\theta  \\ sin\theta & rcos\theta \end{bmatrix}$$

Here $$\underline{A}=\underline{J}^{-1}$$. Therefore it can be written that

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Revision of lecture notes - Explanation of $$\displaystyle \mathbf{A} = \mathbf{J}^{-1}$$

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #006600; background-color: #CCFFFF; text-align: left;"> Sign your comment. The above explanation was not complete; see me after class. Egm6322.s09 00:31, 2 April 2009 (UTC)

Using:

$$\mathbf{A}=\mathbf{J}^{-1}=\frac{\partial x_{i}}{\partial \overline{x}_{j}}=\begin{bmatrix} m & p\\ n & q\end{bmatrix}$$

With the determinant:

$$\frac{1}{det\mathbf{J}^{-1}}\begin{bmatrix}q & -p\\ -n & m\end{bmatrix}\Rightarrow det\mathbf{J}^{-1}=mq-np$$

It is possible to see that:

$$\begin{Bmatrix}\bar{x}\\\bar{y}\end{Bmatrix}=\mathbf{J}\begin{bmatrix} \begin{Bmatrix}x\\y\end{Bmatrix}-\begin{Bmatrix}x_0\\y_0\end{Bmatrix} \end{bmatrix}$$

$$\mathbf{E}=\mathbf{J},\mathbf{F}=-\mathbf{J}\begin{Bmatrix} x_0\\y_0\end{Bmatrix}$$ from pag. 15-1

$$\underline{J} \underline{A}=\underline{I}=\left [\delta _{ij} \right ]=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$$

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Wrong notation $$\displaystyle \partial_{ij}$$; review lecture notes. Also the explanation for $$\displaystyle \underline{A} = \underline{J}^{-1}$$ is incomplete; review lecture notes. Egm6322.s09 17:11, 23 February 2009 (UTC)

$$\underline{J} $$ can easily be determined.

$$\underline{J}= \underline{A}^{-1}=\left (\underline{J}^{-1} \right )^{-1}=\begin{bmatrix} cos\theta & -rsin\theta \\ sin\theta & rcos\theta \end{bmatrix}^{-1}$$

$$\underline{A}^{-1}=\frac{1}{det\underline{A}}\begin{bmatrix} rcos\theta & rsin\theta \\ -sin\theta & cos\theta \end{bmatrix}$$

Eventually,

$$\underline{J}=\frac{1}{r}\begin{bmatrix} rcos\theta & rsin\theta \\ -sin\theta & cos\theta \end{bmatrix}$$

which is identical with the previous result.

=Complete Form Of Matrix=

--EGM6322.S09.TIAN 17:38, 24 April 2009 (UTC)

$$\alpha = \beta + \gamma$$

$$\beta$$ = $$\left \lfloor \partial_{\bar{x}} \; \partial_{\bar{y}} \right \rfloor $$ $$\mathbf{J} \mathbf{J^T}\begin{Bmatrix}\partial {\bar{x}}\\\partial {\bar{y}}\end{Bmatrix}$$

$$= \left \lfloor \partial_r \; \partial_{\theta} \right \rfloor $$ $$\frac {1} {r^2} $$ $$\begin{bmatrix} rc & rs\\ -s & c \end{bmatrix} $$ $$\begin{bmatrix} rc & -s\\ rs & c \end{bmatrix} $$ $$\begin{Bmatrix}\partial_r \\\ \partial_{\theta}\end{Bmatrix}$$

$$= \left \lfloor \partial_r \ \partial_{\theta} \right \rfloor $$ $$\begin{Bmatrix}\partial_r \\\ \partial_{\theta}\end{Bmatrix}$$

$$= \partial_{rr} +$$ $$\frac {1} {r^2} $$ $$\partial_{\theta \theta} $$

Note: $$\left \lfloor \partial_{\bar{x}} \ \partial_{\bar{y}} \right \rfloor $$$$\mathbf{J} \mathbf{J^T}$$ hold fixed, no further differentiate.

Three Methods To Figure Out $$\gamma$$
--EGM6322.S09.TIAN 17:39, 24 April 2009 (UTC)

Method One:

The Jacobian Matrix is:

$$J_{11}=\frac{x} {r}=\frac{rcos\Theta }{r}=cos\Theta$$

$$J_{12}=\frac{y}{\left (x^{2}+y^{2} \right )^{1/2}}=\frac{rsin\theta }{r}=sin\theta $$

$$J_{21}=\frac{1}{1+\left (\frac{y}{x} \right )^{2}}\left (-\frac{y}{x^{2}} \right )=\frac{-y}{x^{2}+y^{2}}=\frac{-rsin\theta }{r^{2}}=-\frac{sin\theta }{r}$$

$$J_{22}=\frac{\partial \overline{x}_{2}}{\partial x_{2}}=\frac{\partial \theta }{\partial y}$$

$$ \underline J=[J_{ij}]=[\frac {\partial {\bar {x_i}}}{\partial {x_j}}] =\begin{bmatrix} c & s     \\ -\frac {s}{r} & \frac{c}{r} \end{bmatrix} =\frac{1}{r}\begin{bmatrix} rc & rs     \\ -s & c \end{bmatrix} $$

$$\gamma =$$ $$= \left \lfloor \partial_r \ \partial_{\theta} \right \rfloor $$ $$\mathbf{J} \mathbf{J^T}$$ $$\begin{Bmatrix}\partial r \\\ \partial {\theta}\end{Bmatrix}$$

Method Two:

Get $$\underline {J}$$ directly in $$(r,\theta)$$ (polar coordinates)

$$x_{i}=x_{i}\left (\overline{x}_{1},\overline{x}_{2} \right )=x_{i}\left (r,\theta \right )$$

$$\begin{bmatrix} \frac{\partial x_{i}}{\partial \overline{x}_{j}} \end{bmatrix} =\begin{bmatrix} cos\theta &-rsin\theta  \\ sin\theta & rcos\theta \end{bmatrix}$$

$$\underline{A}=\begin{bmatrix} \frac{\partial x_{i}}{\partial \overline{x}_{j}} \end{bmatrix} =\begin{bmatrix} cos\theta &-rsin\theta  \\ sin\theta & rcos\theta \end{bmatrix}$$

$$\underline{A}=\underline{J}^{-1}=\frac{\partial x_{i}}{\partial \overline{x}_{j}} =\begin{bmatrix} a & b\\ c & d \end{bmatrix}$$

Therefore,

$$\frac{1}{det\mathbf{J}^{-1}}\begin{bmatrix}d & -b\\ -c & a\end{bmatrix}\Rightarrow det\mathbf{J}^{-1}=1$$

$$\underline{J}=\frac{1}{r}\begin{bmatrix} rcos\theta & rsin\theta \\ -sin\theta & cos\theta \end{bmatrix}=\frac{1}{r}\begin{bmatrix} rc & rs     \\ -s & c \end{bmatrix}$$

Therefore,

$$\gamma= \left \lfloor \partial_x \ \partial_y \right \rfloor $$ $$\mathbf{J} (x,y)$$ $$\begin{Bmatrix}\partial r \\\ \partial {\theta}\end{Bmatrix}$$

Method Three:

re-expect result in $$( r, $$ $$\theta$$ )

$$= \left \lfloor \partial_x \ \partial_y \right \rfloor $$ $$\mathbf{J} (r,\theta)$$ $$\begin{Bmatrix}\partial_r \\\ \partial_{\theta}\end{Bmatrix}$$

expressing result in $$( r, $$ $$\theta$$ )

Let us use Method Three:

$$= \left \lfloor \partial_x \ \partial_y \right \rfloor $$ $$\mathbf{J^T}$$ = $$= \left \lfloor \partial_x \ \partial_y \right \rfloor $$ $$\begin{bmatrix} c & -\frac {s}{r}\\ s & \frac {c}{s} \end{bmatrix} $$ = $$= \left \lfloor ( \partial_x c + \partial_y s) \;  (-\partial_x (\frac {s}{c})) + \partial_y (\frac {c}{r}) \right \rfloor $$

$$\partial_x c =$$ $$\frac{\partial}{\partial x}$$ $$cos {\theta}(x,y)$$ = $$\frac{\partial}{\partial r}$$ $$( cos {\theta})$$ $$\frac{\partial r}{\partial x}$$ + $$\frac{\partial}{\partial {\theta}}$$ $$( cos {\theta})$$ $$\frac{\partial {\theta}}{\partial x}$$

$$\partial_x f(r,{\theta})$$ = ($$\frac{\partial}{\partial r} \; f$$) $$\frac{\partial r}{\partial x} $$ + $$\frac{\partial f}{\partial {\theta}} $$ $$\frac{\partial {\theta}}{\partial x} $$

$$\frac{\partial {\theta}}{\partial x} $$ = $$\frac{\partial}{\partial x} $$ $$tan^{-1}$$ $$( \frac{y}{x} )$$ = $$J_{21} (r,{\theta})$$ = $$\frac{\partial {\bar{x_2}}}{\partial {x_1}} $$

$$J_{21}=$$ $$-\frac {sin{\theta}} {r}$$

$$\partial_x cos {\theta} $$

= $$(-sin{\theta})$$ $$(-\frac {s}{r})$$ = $$\frac {s^2}{r}$$

$$\partial_y sin {\theta}$$ = $$\frac {c^2}{r}$$

$$\Rightarrow$$ $$\partial_x c + \partial_y s$$ = $$\frac {1}{r}$$

$$\partial_x (\frac {s}{r})=$$ $$\frac{\partial} {\partial {r}} $$ $$(\frac {s}{r})$$ $$\frac{\partial r }{\partial x} +$$ $$\frac{\partial} {\partial {\theta}} $$ $$(\frac {s}{r})$$ $$\frac{\partial {\theta} }{\partial x} $$

= $$-\frac {2sc} {r^2}$$

Similarly, $$\partial_y (\frac {c}{r})=$$ $$\frac{\partial} {\partial {r}} $$ $$(\frac {c}{r})$$ $$\frac{\partial r }{\partial y} +$$ $$\frac{\partial} {\partial {\theta}} $$ $$(\frac {c}{r})$$ $$\frac{\partial {\theta} }{\partial y} $$

= $$-\frac {2sc} {r^2}$$

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> But you did not do Method 1 and Method 2. Also, it is not clear to me why you need to box "The complete form of $$\gamma$$" below, and the expression of $$\gamma$$ is wrong. The expression for $$\gamma$$ here is not the same expression for $$\gamma$$ in the section on Example, Rotation and Translation. You could have completed this topic, even though I only explained in words, without writing down. I did write down the end of the complete solution at the beginning of the following lecture. Egm6322.s09 17:11, 23 February 2009 (UTC)

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #006600; background-color: #CCFFFF; text-align: left;"> You still haven't completed Method 1 and Method 2; there were also a lot of "misprints" in the above text around Method 1 and Method 2; clean up.

You deleted the collapsible box "The complete form of $$\gamma$$" but did not reply in the comment box above. Egm6322.s09 00:31, 2 April 2009 (UTC)

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #006600; background-color: #CCFFFF; text-align: left;"> This part has been changed. Egm6322.s09.three.liu 15:20, 10 April 2009 (UTC)

= References =

= Signatures = Egm6322.s09.three.liu 19:02, 25 February 2009 (UTC)

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> You need to sign with the "four tilde" command so you have the time-and-date stamp, just like everybody else; see below. Egm6322.s09 17:11, 23 February 2009 (UTC)

Egm6322.s09.Three.nav 20:37, 19 February 2009 (UTC)Egm6322.s09.three.nav

--Egm6322.s09.xyz 17:52, 18 February 2009 (UTC)

Egm6322.s09.bit.la 16:36, 20 February 2009 (UTC)

Egm6322.s09.bit.sahin 18:26, 20 February 2009 (UTC)

--Egm6322.s09.lapetina 19:12, 20 February 2009 (UTC)

Egm6322.s09.Three.ge 19:23, 20 February 2009 (UTC)

Egm6322.s09.bit.gk 20:05, 20 February 2009 (UTC)

--EGM6322.S09.TIAN 21:23, 20 February 2009 (UTC)

title: User:Egm6322.s09.mafia/HW3


text:
Assistant teacher program/Developing countries/Team sports and sports mentoring

Team sports
Team sports allow to train mediation, teamwork, collective intelligence, higher-order volitions, democracy, mentoring, understanding for the categorical imperative, observing rules and logical understanding for complicated rules, all of this in a context that is easily accessible and motivating for young children. These educational objectives must be made explicit aims, otherwise the learning effect may be marginal. In developing countries it may be especially interesting to make use of this context and its accessibility because accessibility in purely educational classroom situations may be more limited. The context can also help to motivate the necessary attitudes for future assistant teachers. A player who can be a good team captain may have some of the necessary skills and attitudes for an assistant teacher.

Sports mentoring
Sports mentoring is easy and very accessible, at first the pupils do not have to train any attitude other than willingness to help an assigned protégé to improve his personal skills. One could say every sport mentor is a team captain for the smallest possible team. In order to qualify as a sport mentor a pupil should have at least an average grade in the subject sport, he should have shown a degree of understanding for theory (e.g. first aid, physiology and game rules) and should be willing to make a serious attempt to be a good mentor. It is important that the mentor trainers or teachers understand the educational objectives outside the subject sport and direct the training and development of sport mentors towards these goals. Of course that should not lead towards rejecting sport as entertainment and as an end in itself.


text:
User:Wolf3587

This is a little information about the user "Wolf3587" (real name Nicholas Grace).

I currently live in Solihull (just outside Birmingham) in the UK, but have travelled widely in my life. I have travelled around the world 5 times, lived in 17 different countries and visited over 50 different countries.

I currently describe myself as the rather non-PC desription of "Cripple" being that I am disabled person / wheelchair user due to an ongoing abdominal pain problem, which was undiagnosed from 2000 to late 2007, at which point I was given a possible diagnosis of "Visceral Hyperpathia" (don't worry if you have never heard of it, as none of the doctors I come across, day-to-day have either (in fact, the only person that seems to have even heard of it is the specialist that gave the possible diagnosis). This pain means that I am unable to walk more than 3-4 steps without collapsing into unconsciousness witht he pain - obviously, not a lot of fun for someone who has spent much of his life travelling.

This disability has also effected me in other ways besides the obvious physical ones. It has left me socially isolated (unable to get out and about much), gradual and constant decline on my abilities to concentrate, use my former vocabulary and love of learning. I know that my IQ was measured, just before I fell ill, at around 154, but am confident that this is much lower now (and feels as if it drops at least a point a day). Although I have always had a lifelong love of learning, my ability to do so has been severly reduced over the past few years (being unable to concentrate as well as I used to, unable to get to and from places of learning or groups or societies to discuss and learn and having to live on an extremely low income which seriously curtails my ability to study formally).

Professionally, I have had a fairly varied career too. I was originally accepted to study Medicine at university, but had to drop out after the death of my father (who actually died at the time of my O-Levels, but the continued family breakdown left me unable to commit fully to my studies) and my need to bring up my own children. I then more or less alternated between a few professions at the same time: Running public houses - a job I loved to do, but did not pay much; working as an accountant - a job that paid well, but I did not find as enjoyable; running two businesses (one computer company and the other a management consultancy / book-keeping business), which provided me with some extra financial security to allow me to do other things I wanted to do; and a professional musician, touring with a couple of symphony orchestras, jazz bands, blues bands and writing for and performing with a couple of rock / punk bands. A combination of these jobs allowed me to move around and travel as much as I did

title: User:Wolf3587


text:
User:Kifaro50

Hello!

I am a graduate student at the University of Florida currently pursuing a master's degree in Interdisciplinary Ecology. I work in southern Africa studying impacts of African elephants on large mammals and vegetation. I have strong interests in animals and in spending time in the outdoors camping, hiking, and backpacking.

Thanks for visiting!

title: User:Kifaro50


text: Survey design/References

title: Survey design/References


text:
User:JamesSonger

1 bio 2 assignments 2.1 midterm 2.2 final 2.3 in-class writing 3 reflective assessment

title: User:JamesSonger


text:
User:Charitwo/monobook.css

@import url('http://meta.wikimedia.org/w/index.php?title=User:Charitwo/global.css&action=raw&ctype=text/css');

title: User:Charitwo/monobook.css


text:
User:Angelkeziah

Salut! I'm Kei.

title: User:Angelkeziah


text:
User talk:KenzieLockhart

Hello Former Student!

I'm writing to you because you took a class from me in the past few years at the Art Institute. As of Spring, 2010, I'll not be teaching further classes at AIS for the foreseeable future. (We're parting ways on good terms!) Until now, you've been able to reach my Wikiversity educational resources from my website, but I've decided to delete that link. You can still reach all of those materials here.

If you're interested in what I'm up to in the future (which will include an album in 2010 and a novel in 2011) you can friend my facebook page or drop me an email at info@stevenarntson.info and you'll receive a periodic update about performances, classes, and publications.

If you know any current or former students of mine who might be interested in any of this, please forward this information if you have a moment.

I hope all is well for you.
 * Best! Stevenarntson

Personal Essay
Please post your personal essay, and let me know with an email--I'll take a look at it and offer any suggestions I might have.
 * Best, Stevenarntson

title: User talk:KenzieLockhart