text:
Nonlinear finite elements/Calculus of variations

Ideas from the calculus of variations are commonly found in papers dealing with the finite element method. This handout discusses some of the basic notations and concepts of variational calculus. Most of the examples are from  Variational Methods in Mechanics by T. Mura and T. Koya, Oxford University Press, 1992.

The calculus of variations is a sort of generalization of the calculus that you all know. The goal of variational calculus is to find the curve or surface that minimizes a given function. This function is usually a function of other functions and is also called a  functional.

Maxima and minima of functions
The calculus of variations extends the ideas of maxima and minima of functions to functionals.

For a function of one variable $$f(x)$$, the minimum occurs at some point $$x_{\text{min}}$$. For a functional, instead of a point minimum, we think in terms of a function that minimizes the functional. Thus, for a functional $$I[f(x)]$$ we can have a minimizing function $$f_{\text{min}}(x)$$.

The problem of finding  extrema (minima and maxima) or points of inflection (saddle points) can either be  constrained or  unconstrained.

The unconstrained problem.
Suppose $$f(x)$$ is a function of one variable. We want to find the maxima, minima, and points of inflection for this function. No additional constraints are imposed on the function. Then, from elementary calculus, the function $$f(x)$$ has Any point where the condition $$\cfrac{df}{dx} = 0$$ is satisfied is called a  stationary point and we say that the function is stationary at that point.
 * a minimum if $$\cfrac{df}{dx} = 0$$ and $$\cfrac{d^2 f}{dx^2} > 0$$.
 * a maximum if $$\cfrac{df}{dx} = 0$$ and $$\cfrac{d^2 f}{dx^2} < 0$$.
 * a point of inflection if $$\cfrac{d^2 f}{dx^2} = 0$$.

A similar concept is used when the function is of the form $$f(x_1,x_2,x_3,t)$$. Then, the function $$f$$ is stationary if

df = \frac{\partial f}{\partial x_1} dx_1 + \frac{\partial f}{\partial x_2} dx_2 + \frac{\partial f}{\partial x_3} dx_3 + \frac{\partial f}{\partial t} dt = 0~. $$ Since $$x_1$$, $$x_2$$, $$x_3$$, and $$t$$ are  independent variables, we can write the stationarity condition as

\frac{\partial f}{\partial x_1} = 0 ~; \frac{\partial f}{\partial x_2} = 0 ~; \frac{\partial f}{\partial x_3} = 0 ~; \frac{\partial f}{\partial t} = 0~. $$

The constrained problem - Lagrange multipliers.
Suppose we have a function $$f(x_1,x_2,x_3)$$. We want to find the minimum (or maximum) of the function $$f$$ with the added constraint that
 * $$\text{(1)} \qquad

g(x_1,x_2,x_3) = 0 ~. $$ The added constraint is equivalent to saying that the variables $$x_1$$, $$x_2$$, and $$x_3$$ are  not independent and we can write one of the variables in terms of the other two. The stationarity condition for $$f$$ is
 * $$\text{(2)} \qquad

df = \frac{\partial f}{\partial x_1} dx_1 + \frac{\partial f}{\partial x_2} dx_2 + \frac{\partial f}{\partial x_3} dx_3 = 0~.  $$ Since the variables $$x_1$$, $$x_2$$, and $$x_3$$ are not independent, the coefficients of $$dx_1$$, $$dx_2$$, and $$dx_3$$ are not zero.

At this stage we could express $$x_3$$ in terms of $$x_1$$ and $$x_2$$ using the constraint equation (1), form another stationarity condition involving only $$x_1$$ and $$x_2$$, and set the coefficients of $$dx_1$$ and $$dx_2$$ to zero. However, it is usually impossible to solve equation (1) analytically for $$x_3$$. Hence, we use a more convenient approach called the  Lagrange multiplier method.

Lagrange multiplier method.
From equation (1) we have

dg = \frac{\partial g}{\partial x_1} dx_1 + \frac{\partial g}{\partial x_2} dx_2 + \frac{\partial g}{\partial x_3} dx_3 = 0~.  $$ We introduce a parameter $$\lambda$$ called the  Lagrange multiplier and using equation (2) we get

df + \lambda dg = 0~. $$ Then we have,

\left(\frac{\partial f}{\partial x_1} + \lambda\frac{\partial g}{\partial x_1}\right) dx_1 + \left(\frac{\partial f}{\partial x_2} + \lambda\frac{\partial g}{\partial x_2}\right) dx_2 + \left(\frac{\partial f}{\partial x_3} + \lambda\frac{\partial g}{\partial x_3}\right) dx_3 = 0~.  $$ We choose the parameter $$\lambda$$ such that
 * $$\text{(3)} \qquad

\frac{\partial f}{\partial x_3} + \lambda\frac{\partial g}{\partial x_3} = 0~. $$ Then, because $$x_1$$ and $$x_2$$ are independent, we must have
 * $$\text{(4)} \qquad

\frac{\partial f}{\partial x_1} + \lambda\frac{\partial g}{\partial x_1} = 0 \text{and} \frac{\partial f}{\partial x_2} + \lambda\frac{\partial g}{\partial x_2} = 0 $$ We can now use equations (1), (3), and (4) to solve for the extremum point and the Lagrange multiplier. The constraint is satisfied in the process.

Notice that equations (1), (3) and (4) can also be written as

{   \frac{\partial h}{\partial \lambda} = 0 ~; \frac{\partial h}{\partial x_1} = 0 ~; \frac{\partial h}{\partial x_2} = 0 ~; \frac{\partial h}{\partial x_3} = 0 } $$ where

{   h(x_1, x_2, x_3, \lambda) := f(x_1,x_2,x_3) + \lambda g(x_1,x_2,x_3)~. } $$

Minima of functionals
Consider the functional
 * $$\text{(5)} \qquad

I[y(x)] = \int_{x_0}^{x_1} \left[f(x)\left(\cfrac{dy(x)}{dx}\right)^2 + g(x)y(x)^2 + 2h(x)y(x)\right] ~dx~. $$ We wish to minimize the functional $$I$$ with the constraints (prescribed boundary conditions)

y(x_0) = y_0 ~,~ y(x_1) = y_1~. $$

Let the function $$y = y(x)$$ minimize $$I$$. Let us also choose a  trial function (that is not quite equal to the solution $$y(x)$$)
 * $$\text{(6)} \qquad

y = y(x) + \lambda v(x) $$ where $$\lambda$$ is a parameter, and $$v(x)$$ is an arbitrary continuous function that has the property that

v(x_0) = 0 \text{and} v(x_1) = 0~. $$ (See Figure 1 for a geometric interpretation.)

Plug (6) into (5) to get
 * $$\text{(8)} \qquad

I[y(x)+\lambda v(x)] = \int_{x_0}^{x_1} \left[f(x)\left(\cfrac{dy(x)}{dx}+        \lambda\cfrac{dv}{dx}\right)^2 + g(x)\left[y(x) + \lambda v(x)\right]^2 + 2h(x)\left[y(x) + \lambda v(x)\right]\right] ~dx~. $$ You can show that equation (8) can be written as (show this)

I[y(x)+\lambda v(x)] = I[y(x)] + \delta I + \delta^2 I      ~\text{or,}~ I[y(x)+\lambda v(x)] - I[y(x)] = \delta I + \delta^2 I $$ where
 * $$\text{(9)} \qquad

\delta I = 2\lambda \int_{x_0}^{x_1} \left[f(x) \left(\cfrac{dy(x)}{dx}\right) \left(\cfrac{dv(x)}{dx}\right) + g(x) y(x) v(x) + h(x) v(x)\right]~dx $$ and
 * $$\text{(10)} \qquad

\delta^2 I = \lambda^2 \int_{x_0}^{x_1} \left[f(x)\left(\cfrac{dv(x)}{dx}\right)^2 + g(x)[v(x)]^2\right]~dx~. $$

The quantity $$\delta I$$ is called the  first variation of $$I$$ and the quantity $$\delta^2 I$$ is called the  second variation of $$I$$. Notice that $$\delta I$$ consists only of terms containing $$\lambda$$ while $$\delta^2 I$$ consists only of terms containing $$\lambda^2$$.

The  necessary condition for $$I[y(x)]$$ to be a minimum is
 * $$\text{(11)} \qquad

{   \delta I = 0 ~. } $$

Remark.
The  first variation of the functional $$I[y]$$ in the direction $$v$$ is defined as

{   \delta I(y;v) = \lim_{\epsilon\rightarrow 0} \cfrac{I[y + \epsilon v] - I[y]}{\epsilon} \equiv \left.\cfrac{d}{d\epsilon} I[y + \epsilon v]\right|_{\epsilon = 0} ~.} $$

To find which function makes $$\delta I$$ zero, we first integrate the first term of equation (9) by parts. We have,

\int_{x_0}^{x_1} \left(f \cfrac{dy}{dx}\right) \cfrac{dv}{dx}~dx = \left[\left(f \cfrac{dy}{dx}\right) v \right]_{x_0}^{x_1} - \int_{x_0}^{x_1} \cfrac{d}{dx}\left(f \cfrac{dy}{dx}\right) v~dx~. $$ Since $$v = 0$$ at $$x_0$$ and $$x_1$$, we have
 * $$\text{(12)} \qquad

\int_{x_0}^{x_1} \left(f \cfrac{dy}{dx}\right) \cfrac{dv}{dx}~dx = - \int_{x_0}^{x_1} \cfrac{d}{dx}\left(f \cfrac{dy}{dx}\right) v~dx $$ Plugging equation (12) into (9) and applying the minimizing condition (11), we get

0 = \int_{x_0}^{x_1} \left[-\cfrac{d}{dx}\left(f(x)           \cfrac{dy(x)}{dx}\right) v(x) + g(x) y(x) v(x) + h(x) v(x)\right]~dx $$ or,
 * $$\text{(13)} \qquad

\int_{x_0}^{x_1} \left[-\cfrac{d}{dx}\left(f(x)           \cfrac{dy(x)}{dx}\right) + g(x) y(x) + h(x) \right]v(x)~dx = 0~. $$

The  fundamental lemma of variational calculus states that if $$u(x)$$ is a piecewise continuous function of $$x$$ and $$v(x)$$ is a continuous function that vanishes on the boundary, then
 * $$\text{(14)} \qquad

{   \int_{x_0}^{x_1} u(x) v(x) ~dx = 0 \implies u(x) = 0 ~. } $$ Applying (14) to (13) we get
 * $$\text{(15)} \qquad

-\cfrac{d}{dx}\left(f(x)\cfrac{dy(x)}{dx}\right) + g(x) y(x) + h(x) = 0~. $$ Equation (15) is called the  Euler equation of the functional $$I$$. The solution of the Euler equation is the minimizing function that we seek.

Of course, we cannot be sure that the solution represents and minimum unless we check the second variation $$\delta^2 I$$. From equation (10) we can see that $$\delta^2 I > 0$$ if $$f(x) > 0$$ and $$g(x) > 0$$ and in that case the problem is guaranteed to be a minimization problem.

We often define

\delta y := \lambda v(x) \text{and} \delta y^{'} := \lambda \cfrac{dv(x)}{dx} $$ where $$\delta y$$ is called a  variation of $$y(x)$$.

In this notation, equation (9) can be written as

\delta I = 2\int_{x_0}^{x_1}\left[f\left(\cfrac{dy}{dx}\right)\delta y^{'} + g y \delta y + h \delta y\right]~dx $$ You see this notation in the principle of virtual work in the mechanics of materials.

An example
Consider the string of length $$l$$ under a tension $$T$$ (see Figure 2). When a vertical load $$f$$ is applied, the string deforms by an amount $$u(x)$$ in the $$y$$-direction. The deformed length of an element $$dx$$ of the string is

ds = \sqrt{1 + \left(\cfrac{du}{dx}\right)^2} dx ~. $$ If the deformation is small, we can expand the relation into a Taylor series and ignore the higher order terms to get

ds = \left[1 + \frac{1}{2}\left(\cfrac{du}{dx}\right)^2\right] dx ~. $$ The force T in the string moves a distance

ds - dx = \frac{1}{2}\left(\cfrac{du}{dx}\right)^2~dx. $$ Therefore, the work done by the force $$T$$ (per unit original length of the string) (the stored elastic energy) is

\frac{1}{2} T \left(\cfrac{du}{dx}\right)^2~. $$ The work done by the forces $$f$$ (per unit original length of string) is

f u $$ We want to minimize the total energy. Therefore, the functional to be minimized is

I[y] = \cfrac{T}{2} \int_0^l \left(\cfrac{du}{dx}\right)^2~dx - \int_0^l f u~dx ~. $$ The Euler equation is

T\cfrac{d^2 u}{dx^2} + f = 0~. $$ The solution is

u = \cfrac{f}{2T} (l - x) x~. $$


text: File:StringBend.png

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

title: File:StringBend.png


text: File:TrialFunction.png

title: File:TrialFunction.png


text:
File:Film school FrameForge Studio dots.png

Used with the WikiU Film School Lesson on how to create 3D storyboards with FrameForge 3D Studio Demo Version.

Summary
[[Category:Filmmaking]


text:
User:Mchua

Hi, I'm Mel. I'm experimenting with making Wikiversity my personal learning portal. I'm most active on the OLPC wiki but also hang out (much less actively) on en-Wikipedia and Appropedia.

Things I'm interested in
Short answer: everything.

"More Proper" comma-delimited list answer: Engineering (particularly electrical engineering and computer science) education, learning within hacker/wiki culture, and the intersection of the two; mathetics, autodidacticism and self-directed learning, polymaths/consilience, and interconnections between everything mentioned so far.

Long answer: User:Mchua/Interests

Currently reading

 * Vital Ideation - I'm "taking" this class as the only non-Olin-College-student participant, since they've made it a Wikiversity course as well.
 * Making Wikiversity a personal learning space - what I'm trying to move towards as an experiment; one of my biggest blockers is the difficulty of syncing so many different notes across wikis.
 * Composing free and open online educational resources should read...

Learning sandboxes

 * /Python - my sandbox for learning about Python (I already program).
 * /Trac - my sandbox for learning about Trac.
 * /Poi
 * /Vocal percussion
 * How to design programs (my workpage: /How to design programs)

Stalled projects

 * User:Mchua/Open education
 * User:Mchua/Mandarin through software
 * /Braindumps
 * /Programming - mostly various languages I would like to learn, plus experiments.
 * Ivan Illich: Deschooling Society - I want to spend a day just reading this sometime.

title: User:Mchua


text:
User:Alihadizadeh

Hi ali

title: User:Alihadizadeh


text:
Nonlinear finite elements/Axially loaded bar

An axially loaded bar
Let us consider the simplest solid mechanics problem that we can think of - the axial loading of a bar (see Figure 1).

 Given:

The bar is of length $$L\,$$ and has an area of cross-section $$A\,$$. The Young's modulus of the material is $$E\,$$.

 Find:

We want to find the  stresses and  deformations in the bar due to the concentrated axial load $$\mathbf{R}$$ at the end.

 Assumptions:
 * 1) The cross-section of the bar does not change after loading.
 * 2) The material is linear elastic, isotropic, and homogeneous.
 * 3) The load is centric.
 * 4) End-effects are not of interest to us.

Strength of Materials Approach
From the free-body diagram of the bar, a balance of forces gives

\mathbf{f}_{\text{wall}} = \mathbf{R} = \mathbf{f}(\mathbf{x}) ~. $$

The  stress in the bar is given by

\boldsymbol{\sigma}(\mathbf{x}) := \cfrac{\mathbf{f}(\mathbf{x})}{A} \qquad \implies \qquad {\boldsymbol{\sigma}(\mathbf{x}) = \cfrac{\mathbf{R}}{A}} ~. $$

The  constitutive model for the bar is given by Hooke's law ($$\boldsymbol{\sigma} = E\boldsymbol{\varepsilon}$$). Therefore, the  strain in the bar is

\boldsymbol{\varepsilon}(\mathbf{x}) = \cfrac{\boldsymbol{\sigma}(\mathbf{x})}{E} \qquad \implies \qquad {\boldsymbol{\varepsilon}(\mathbf{x}) = \cfrac{\mathbf{R}}{AE} }~. $$ To get the  deformation of the bar, we use the strain-displacement relations

\boldsymbol{\varepsilon} := \cfrac{\Delta l}{l} = \cfrac{\delta_{(\mathbf{x} = L)}}{L} \qquad \implies \qquad {\delta_{(\mathbf{x} = L)} = \cfrac{\mathbf{R} L}{A E} }~. $$ We can get the deformation of any point in the bar in a similar manner:

\boldsymbol{\varepsilon}(\mathbf{x}) = \cfrac{\delta(\mathbf{x})}{\mathbf{x}} \qquad \implies \qquad {\delta(\mathbf{x}) = \cfrac{\mathbf{R} \mathbf{x}}{A E} }~. $$

A more complicated axial load
Now, suppose that we have a linear distributed axial load $$\mathbf{q}(\mathbf{x}) = a \mathbf{x}$$ in addition to the load $$\mathbf{R}$$ (see Figure 2(a)). This load has units of force per unit length and is also called a  body load. Examples are gravity and inertial forces due to rotation around the $$y$$ axis.

Once again, we want to calculate stresses and displacements. Note that stresses are no longer constant along the length of the bar.

Strength of Materials Approach
From the free body diagram of the bar, a balance of forces gives

\mathbf{f}_{\text{wall}} = \mathbf{R} + \int_0^L \mathbf{q}(\mathbf{x})~dx \qquad \implies \qquad {\mathbf{f}_{\text{wall}} = \mathbf{R} + \cfrac{a L^2}{2}} ~. $$

So we know the reaction force at the wall. But we want to know the stress and displacement at any point along the bar.

To do this, we can make a cut at any point along the length and use the free body diagram to compute a reaction force $$\mathbf{f}(\mathbf{x})$$ (see Figure 2(b)). But this is tedious since we have to repeat the process for every point in the bar.


text: Category:Nonlinear finite elements

title: Category:Nonlinear finite elements


text: File:AxialBar.png

title: File:AxialBar.png


text:
User:Magosgruss

Hey, I am Magosgruss, my interest is in Thelema and general metaphysics.

Tschus!

title: User:Magosgruss


text: File:DistAxialBar.png

title: File:DistAxialBar.png


text: File:DistAxialBarFBD.png

title: File:DistAxialBarFBD.png


text:
Nonlinear finite elements/Axial bar strong form

Axially loaded bar : Strong form
In this case, we consider an  infinitesimal slice of the bar and perform a balance of forces on that slice (see Figure 1).

A balance of forces gives

A \boldsymbol{\sigma} = \mathbf{q}(\mathbf{x})~\Delta x + A (\boldsymbol{\sigma} + \Delta\boldsymbol{\sigma}) \qquad \implies \qquad A \left(\cfrac{(\boldsymbol{\sigma}+\Delta\boldsymbol{\sigma}) - \boldsymbol{\sigma}}{\Delta x}\right) + \mathbf{q}(\mathbf{x}) = 0 ~. $$

Taking the limit as $$\Delta x \rightarrow 0$$, we get

A \lim_{\Delta x \rightarrow 0} \left(\cfrac{(\boldsymbol{\sigma}+\Delta\boldsymbol{\sigma}) - \boldsymbol{\sigma}}{\Delta x}\right) + \mathbf{q}(\mathbf{x}) = 0 \qquad \implies \qquad {    A \cfrac{d\boldsymbol{\sigma}}{dx} + \mathbf{q}(\mathbf{x}) = 0 ~. } $$

This is a differential equation written in terms of stress. We want to express it in terms of displacements. How do we do this?

First, we convert the stresses into strains using the constitutive equation ($$\boldsymbol{\sigma} = E \boldsymbol{\varepsilon}$$).

{    AE \cfrac{d\boldsymbol{\varepsilon}}{dx} + \mathbf{q}(\mathbf{x}) = 0 ~. } $$

Then, we convert the strains into displacements using the  strain-displacement relations:

\boldsymbol{\varepsilon} := \lim_{l\rightarrow 0}\cfrac{\Delta l}{l} \qquad \implies \qquad \boldsymbol{\varepsilon} = \lim_{\Delta x \rightarrow 0} \cfrac{(\mathbf{u} + \Delta\mathbf{u}) - \mathbf{u}}{\Delta x}    \equiv {\boldsymbol{\varepsilon} = \cfrac{d\mathbf{u}}{dx}} ~. $$

The differential equation in terms of the displacement $$\mathbf{u}$$ is

{    AE \cfrac{d^2\mathbf{u}}{dx^2} + \mathbf{q}(\mathbf{x}) = 0 ~. } $$ Since $$\mathbf{q}(\mathbf{x}) = a \mathbf{x}$$, we have an  inhomogeneous ordinary differential equation for the displacements in the bar.

{    AE \cfrac{d^2\mathbf{u}}{dx^2} + a \mathbf{x} = 0 ~. } $$

If we can find the displacements in the bar, then we can solve for the strains and therefore the stresses. But to do that, we have to solve the  governing differential equation.

To get a unique solution, we need to provide  boundary conditions.

In this case, these are Since the ODE is in terms of $$\mathbf{u}$$, the BCs must also be in terms of $$\mathbf{u}$$. But we have a force at one end. This force has to be converted into an equivalent displacement. How?
 * 1) At $$\mathbf{x} = 0$$ (wall), $$\mathbf{u} = 0$$. This is an  essential BC.
 * 2) At $$\mathbf{x} = L$$ (end), $$\mathbf{f} = \mathbf{R}$$. This is a  natural BC.

\mathbf{f} = A \boldsymbol{\sigma} = A E \boldsymbol{\varepsilon} = A E \cfrac{d\mathbf{u}}{dx} ~. $$

Therefore, the BCs become

The problem can then be stated as
 * 1) At $$\mathbf{x} = 0$$ (wall), $$\mathbf{u} = 0$$.
 * 2) At $$\mathbf{x} = L$$ (end), $$A E \cfrac{d\mathbf{u}}{dx} = \mathbf{R}$$.

$$   \begin{align} \text{Find}~ & \mathbf{u}(\mathbf{x})~\text{such that it satisfies} \\ & AE \cfrac{d^2\mathbf{u}}{dx^2} = - a \mathbf{x} \\ & \mathbf{u}(0) = 0 \\ & \left.\cfrac{d\mathbf{u}}{dx}\right|_{\mathbf{x} = L} = \cfrac{\mathbf{R}}{AE} \end{align} $$

The differential formulation is also called the  strong form of the problem.

Analytical Solution
The analytical solution strategy is as follows:
 * 1) Set right hand side to zero and solve ( Homogeneous solution).
 * 2) Find  one solution with RHS = $$-a \mathbf{x}$$ ( Particular solution).
 * 3) General solution = homogeneous + particular.
 * 4) Apply BCs.

Homogeneous solution

 * The homogeneous ODE is

AE \cfrac{d^2\mathbf{u}}{dx^2} = 0~. $$
 * Integrate twice to get the homogeneous solution

A E \mathbf{u} = C_1 \mathbf{x} + C_2 \quad \implies \quad \mathbf{u}(\mathbf{x}) = \cfrac{C_1 \mathbf{x} + C_2}{A E} ~. $$

Particular solution

 * The ODE is

AE \cfrac{d^2\mathbf{u}}{dx^2} = -a \mathbf{x} ~. $$
 * Integrate twice to get the particular solution (and assume that the constants of integration are zero)

A E \mathbf{u} = -\cfrac{a \mathbf{x}^3}{6} \quad \implies \quad \mathbf{u}(\mathbf{x}) = -\cfrac{a \mathbf{x}^3}{6 A E} ~. $$

General solution

 * Add the homogeneous and particular parts to get

\mathbf{u}(\mathbf{x}) = \cfrac{-a \mathbf{x}^3 + D_1 \mathbf{x} + D_2}{6AE} ~. $$

Apply BCs

 * At $$\mathbf{x} = 0$$, $$\mathbf{u} = 0$$. Therefore, $$D_2 = 0$$.
 * At $$\mathbf{x} = L$$,

\cfrac{d\mathbf{u}}{dx} = \cfrac{-3aL^2 + D_1}{6AE} = \cfrac{R}{AE} \quad \implies \quad D_1 = 6R + 3aL^2 ~. $$

Solution

 * The displacement field in the bar is

{       \mathbf{u}(\mathbf{x}) = \cfrac{-a \mathbf{x}^3 + (6\mathbf{R} + 3aL^2)\mathbf{x}}{6AE} ~. }     $$
 * The strain in the bar is

{       \boldsymbol{\varepsilon}(\mathbf{x}) = \cfrac{-a \mathbf{x}^2 + (2\mathbf{R} + aL^2)}{2AE} ~. }     $$
 * The stress in the bar is

\boldsymbol{\sigma}(\mathbf{x}) = \cfrac{-a \mathbf{x}^2 + (2\mathbf{R} + aL^2)}{2A} ~. $$ A plot of this solution is shown in Figure 2. All known quantities have been taken to be 1 for the plot.


text: File:DistAxialBarCont.png

title: File:DistAxialBarCont.png


text: File:DistAxialBarExactSol.png

title: File:DistAxialBarExactSol.png


text:
User talk:Mchua

Hello Mchua, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --HappyCamper 23:11, 12 August 2007 (UTC)

Electrical engr education
Welcome Mchua! I was recently focused on the algebra area, since I get enough of EE at school. If you have any ideas in that area, I'd be happy to help. How was Wikimania?--Rayc 03:54, 20 August 2007 (UTC)

re: algebra curriculum
Nope, nothing seems to be happening in that algebra class as far as I can tell.--Rayc 01:54, 2 September 2007 (UTC)

Wikimania pictures
Hi Mchua! I'd be delighted to see those pictures - do you want to upload them to Commons? Or, if you prefer, you can send them via email - cormaggio at gmail dot com. Cheers! Cormaggio talk 20:00, 19 January 2008 (UTC)

 * Hey Mel - did you ever upload these photos? :-) Also, I'm very interested in your hoping to make Wikiversity your personal learning portal - do you have any thoughts on this so far, or any ideas for how to make it more productive for you? Countrymike has also written about this idea on his blog - I'd really like to start a learning/discussion/reading group on how this can be made work... Cormaggio talk 10:17, 28 February 2008 (UTC)

 * I've set up a page at Making Wikiversity a personal learning space, and a section on the Colloquium at: Colloquium. It would be great to hear your comments! Cormaggio talk 12:56, 29 February 2008 (UTC)

Hi!
Hi Mel! Nice to find you around Wikiversity! :) I've joined recently the Composing free and open online educational resources course to learn more about OER and also have the purpose of participating in Wikiversity, so hope to keep her in contact with you... Where are you at the moment? Did you start to travel around already? --Esenabre 03:09, 14 March 2008 (UTC)

Mentat wiki month
Hi Mel! Yes, I noticed that page starting up, but haven't been around over the last week or so to see it develop. Looks really interesting. I'll keep it on my watchlist and peek in regularly - don't think I can sign up for more courses just now... Cormaggio talk 08:08, 9 May 2008 (UTC)

title: User talk:Mchua


text:
Nonlinear finite elements/Axial bar weak form

Axially loaded bar: Weak Form
Instead of deriving the differential equation using a balance of forces on a differential element, we may arrive at the same problem description via a different route -  the principle of virtual work (also called the variational approach).

If we imagine there were forces (virtual forces) inside and outside of the bar, then the virtual work generated by these 'virtual forces' should conserve energy. For the bar, this principle can be stated as

\delta W_{\text{int}} = \delta W_{\text{ext}} + \delta W_{\text{body}}\, $$ where $$\delta W_{\text{int}}\,$$ is the virtual work of the internal forces, $$\delta W_{\text{ext}}\,$$ is the virtual work of the external forces, and $$\delta W_{\text{body}}\,$$ is the virtual work of the body forces.

The virtual internal work is given by

\delta W_{\text{int}} = A \int_0^L \boldsymbol{\sigma}~\delta \boldsymbol{\varepsilon}~dx ~. $$

The virtual work done by the external forces is given by

\delta W_{\text{ext}} = \left. \mathbf{R}~\delta \mathbf{u}\right|_{\mathbf{x} = L} ~. $$

The virtual work done by the body forces is given by

\delta W_{\text{body}} = \int_0^L \mathbf{q}~\delta \mathbf{u}~dx ~. $$

The principle of virtual work for the bar can then be expressed as

A \int_0^L \boldsymbol{\sigma}~\delta \boldsymbol{\varepsilon}~dx = \int_0^L \mathbf{q}~\delta \mathbf{u}~dx + \left. \mathbf{R}~\delta \mathbf{u}\right|_{\mathbf{x} = L} ~. $$

Now, the stress and the virtual strain expressed in terms of the displacements are

\boldsymbol{\sigma} = E \cfrac{d\mathbf{u}}{dx} \qquad \text{and} \qquad \delta \boldsymbol{\varepsilon} = \cfrac{d(\delta \mathbf{u})}{dx} ~. $$

Therefore, we have

A \int_0^L E\cfrac{d\mathbf{u}}{dx}\cfrac{d{\delta \mathbf{u}}}{dx}~dx = \int_0^L \mathbf{q}~\delta \mathbf{u}~dx + \left. \mathbf{R}~\delta \mathbf{u}\right|_{\mathbf{x} = L} ~. $$ Note that the virtual displacement $$\delta \mathbf{u}\,$$ is zero at points on the boundary where displacements are prescribed.

The above equation is called the  variational form or  weak form of the problem.

Why is it called a variational form?
Let us start with the weak form. Using the formula for the first variation (from variational calculus)

\delta (x^2) = 2~x~\delta x \, $$ we have

A \int_0^L E~\cfrac{d\mathbf{u}}{dx}~\cfrac{d\delta\mathbf{u}}{dx}~dx = A \int_0^L \cfrac{E}{2}~\delta \left(\cfrac{d\mathbf{u}}{dx}\right)^2~dx = \delta \left[A \int_0^L \cfrac{E}{2}\left(\cfrac{d\mathbf{u}}{dx}\right)^2~dx \right]~. $$

Therefore, the weak form can be written as

\delta \left[\int_0^L \left(\cfrac{AE}{2}~\left(\cfrac{d\mathbf{u}}{dx}\right)^2 - \mathbf{q}~\mathbf{u}\right)~dx - \left. \mathbf{R}~\mathbf{u}\right|_{\mathbf{x} = L}\right] = 0 ~. $$

This is equivalent to the following  variational statement of the problem $$ \begin{align} \text{Find}~ & \mathbf{u}(\mathbf{x})~\text{such that it satisfies} \\ & \delta I = 0 \\ \text{where}~ & I~\text{is the functional defined as} \\ & I = \int_0^L\cfrac{AE}{2}\left(\cfrac{d\mathbf{u}}{dx}\right)^2~dx - \int_0^L\mathbf{q}~\mathbf{u}~dx - \mathbf{R}\mathbf{u}|_{x = L} \\ \text{with}&\\ & \mathbf{u}(0) = 0 \\ & \delta \mathbf{u}(0) = 0 \end{align} $$

where $$\delta \,$$ means variation in and $$\delta \mathbf{u}\,$$ is an arbitrary variation on $$\mathbf{u}\,$$ subject to the condition that $$\delta \mathbf{u}(0) = 0\,$$.

Principle of minimum potential energy
Note that the first term of $$I$$ is the strain energy stored and the next two terms are the potential energy of the external forces ($$\mathbf{u},\mathbf{R}$$ ). The above variational statement can also be interpreted as statement of minimum potential energy: the total potential of the system (elastic body and external forces) attains a stationary value when equilibrium is satisfied. Further more, this stationary value is minimum. This can be reinterpreted as  of all the possible displacements, the actual displacement $$\mathbf{u}$$ minimizes the potential energy of the system.

Equivalence of the strong and weak forms
The strong and weak statements of the problem are equivalent. Since finite elements uses the  weak form of the problem statement, we can either derive the weak form from the strong form or using the principle of virtual work. The preferred approach is chosen on the basis of convenience.

We can derive the weak form from the strong form as follows. We multiply the strong form with an arbitrary virtual displacement (or weighting function) and integrate over the length of the bar. Thus
 * $$\text{(5)} \qquad

\int_0^L\left(AE \cfrac{d^2\mathbf{u}}{dx^2} + \mathbf{q}\right)\delta \mathbf{u}~dx = 0 ~. $$ Next, we integrate the first term of the equation by parts (to get rid of the higher order derivatives of $$\mathbf{u}$$) and get

\int_0^L AE \cfrac{d\mathbf{u}}{dx}\cfrac{d(\delta \mathbf{u})}{dx}~dx = \int_0^L\mathbf{q}\delta \mathbf{u}~dx + \left. AE\cfrac{d\mathbf{u}}{dx}\delta \mathbf{u}\right|_0^L ~. $$ Recall that $$\delta \mathbf{u}(0) = 0\,$$ and $$\mathbf{R} = AE\cfrac{d\mathbf{u}}{dx}$$. Therefore, we have

\int_0^L AE \cfrac{d\mathbf{u}}{dx}\cfrac{d(\delta \mathbf{u})}{dx}~dx = \int_0^L\mathbf{q}\delta \mathbf{u}~dx + \left. \mathbf{R}~\delta \mathbf{u}\right|_{x=L} ~. $$ This is the same as the weak form derived from the principle of virtual work.


text:
Nonlinear finite elements/Axial bar approximate solution

Approximate Solution: The Galerkin Approach
To find the finite element solution, we can either start with the strong form and derive the weak form, or we can start with a weak form derived from a variational principle.

Let us assume that the approximate solution is $$\mathbf{u}_h(\mathbf{x})$$ and plug it into the ODE. We get

AE \cfrac{d^2\mathbf{u}_h}{dx^2} + a \mathbf{x} = R_h(\mathbf{x}) $$ where $$R_h$$ is the  residual. We now try to minimize the residual in a weighted average sense

\int_0^L R_h(\mathbf{x}) \mathbf{w}(\mathbf{x}) ~dx = 0 $$ where $$\mathbf{w}(\mathbf{x})$$ is a weighting function. Notice that this equation is similar to equation (5) (see 'Weak form: integral equation') with $$\mathbf{w}$$ in place of the variation $$\delta \mathbf{u}$$. For the two equations to be equivalent, the weighting function must also be such that $$\mathbf{w}(0) = 0$$.

Therefore the approximate weak form can be written as

{   \int_0^L AE \cfrac{d\mathbf{u}_h}{dx}\cfrac{d\mathbf{w}}{dx}~dx  = \int_0^L\mathbf{q}\mathbf{w}~dx + \left. \boldsymbol{R}~\mathbf{w}\right|_{x=L} ~. } $$ In Galerkin's method we assume that the approximate solution can be expressed as

\mathbf{u}_h(\mathbf{x}) = a_1\varphi_1(\mathbf{x}) + a_2\varphi_2(\mathbf{x}) + \dots + a_n\varphi_n(\mathbf{x}) = \sum_{i=1}^n a_i \varphi_i(\mathbf{x}) ~. $$ In the  Bubnov-Galerkin method, the weighting function is chosen to be of the same form as the approximate solution (but with arbitrary coefficients),

\mathbf{w}(\mathbf{x}) = b_1\varphi_1(\mathbf{x}) + b_2\varphi_2(\mathbf{x}) + \dots + b_n\varphi_n(\mathbf{x}) = \sum_{j=1}^n b_j \varphi_j(\mathbf{x}) ~. $$ If we plug the approximate solution and the weighting functions into the approximate weak form, we get

\int_0^L AE \left(\sum_{i=1}^n a_i\cfrac{d\varphi_i}{dx}\right) \left(\sum_{j=1}^n b_j\cfrac{d\varphi_j}{dx}\right)~dx = \int_0^L\mathbf{q}\left(\sum_{j=1}^n b_j\varphi_j\right)~dx + \left. \boldsymbol{R}~\left(\sum_{j=1}^n b_j\varphi_j\right)\right|_{x=L} ~. $$ This equation can be rewritten as

\sum_{j=1}^n b_j \left[\int_0^L AE \left(\sum_{i=1}^n a_i\cfrac{d\varphi_i}{dx}                                            \cfrac{d\varphi_j}{dx}\right)~dx \right] = \sum_{j=1}^n b_j \left[\int_0^L\mathbf{q}\varphi_j~dx + \left. \left(\boldsymbol{R}~\varphi_j\right)\right|_{x=L}\right] ~. $$ From the above, since $$b_j$$ is arbitrary, we have

\int_0^L AE \left(\sum_{i=1}^n a_i\cfrac{d\varphi_i}{dx}                                            \cfrac{d\varphi_j}{dx}\right)~dx = \int_0^L\mathbf{q}\varphi_j~dx + \left. \boldsymbol{R}~\varphi_j\right|_{x=L} ~, ~ j = 1\dots n. $$ After reorganizing, we get

\sum_{i=1}^n \left[\int_0^L \cfrac{d\varphi_j}{dx} AE         \cfrac{d\varphi_i}{dx}~dx\right] a_i = \int_0^L \varphi_j\mathbf{q}~dx + \left. \varphi_j\boldsymbol{R}\right|_{x=L} ~, ~ j = 1\dots n $$ which is a system of $$n$$ equations that can be solved for the unknown coefficients $$a_i$$. Once we know the $$a_i$$s, we can use them to compute approximate solution. The above equation can be written in matrix form as

\mathbf{K} \mathbf{a} = \mathbf{f} \qquad \leftrightarrow \qquad K_{ji} a_i = f_j $$ where

\mathbf{K} = \int_0^L \mathbf{B}^T \mathbf{D} \mathbf{B}~ dx \qquad \leftrightarrow \qquad K_{ji} = \int_0^L \cfrac{d\varphi_j}{dx} AE \cfrac{d\varphi_i}{dx}~dx $$ and

f_j = \int_0^L \varphi_j\mathbf{q}~dx + \left. \varphi_j\boldsymbol{R}\right|_{x=L}~. $$

The problem with the general form of the Galerkin method is that the functions $$\varphi_i$$ are difficult to determine for complex domains.


text:
Nonlinear finite elements/Axial bar finite element solution

Axially loaded bar: The Finite Element Solution
The  finite element method is a type of Galerkin method that has the following advantages: As a result, computations can be done in a modular manner that is suitable for computer implementation.
 * 1) The functions $$\varphi_i$$ are found in a systematic manner.
 * 2) The functions $$\varphi_i$$ are chosen such that they can be used for arbitrary domains.
 * 3) The functions $$\varphi_i$$ are piecewise polynomials.
 * 4) The functions $$\varphi_i$$ are non-zero only on a small part of the domain.

Discretization
The first step in the finite element approach is to divide the domain into  elements and  nodes, i.e., to create the finite element mesh.

Let us consider a simple situation and divide the rod into 3 elements and 4 nodes as shown in Figure 6.

Shape functions
The functions $$\varphi_i$$ have special characteristics in finite element methods and are generally written as $$N_i$$ and are called basis functions,  shape functions, or  interpolation functions. Therefore, we may write
 * $$\begin{align}

K_{ij} & = \int_0^L \cfrac{dN_j}{dx} AE \cfrac{dN_i}{dx}~dx \\ f_j & = \int_0^L N_j\mathbf{q}~dx + \left. N_j\boldsymbol{R}\right|_{x=L}~. \end{align}$$

The finite element basis functions are chosen such that they have the following properties:
 * 1) The functions $$N_i$$ are bounded and continuous.
 * 2) If there are $$n$$ nodes, then there are $$n$$ basis functions - one for each node. There are four basis functions for the mesh shown in Figure 6.
 * 3) Each function $$N_i$$ is nonzero only on elements connected to node $$i$$.
 * 4) $$N_i$$ is 1 at node $$i$$ and zero at all other nodes.

Stiffness matrix
Let us compute the values of $$K_{ij}$$ for the three element mesh. We have
 * $$\begin{align}

K_{ij} & = \int_0^L \cfrac{dN_j}{dx} AE \cfrac{dN_i}{dx}~dx \\ & = \int_0^{x_2} \cfrac{dN_j}{dx} AE \cfrac{dN_i}{dx}~dx + \int_{x_2}^{x_3} \cfrac{dN_j}{dx} AE \cfrac{dN_i}{dx}~dx + \int_{x_3}^{L} \cfrac{dN_j}{dx} AE \cfrac{dN_i}{dx}~dx \\ & \equiv \int_{\Omega_1} \cfrac{dN_j}{dx} AE \cfrac{dN_i}{dx}~dx + \int_{\Omega_2} \cfrac{dN_j}{dx} AE \cfrac{dN_i}{dx}~dx + \int_{\Omega_3} \cfrac{dN_j}{dx} AE \cfrac{dN_i}{dx}~dx \end{align}$$ The components of $$\mathbf{K}$$ are
 * $$\begin{align}

K_{11} & = \int_{\Omega_1} \cfrac{dN_1}{dx} AE \cfrac{dN_1}{dx}~dx + \int_{\Omega_2} \cfrac{dN_1}{dx} AE \cfrac{dN_1}{dx}~dx + \int_{\Omega_3} \cfrac{dN_1}{dx} AE \cfrac{dN_1}{dx}~dx \\ K_{12} & = \int_{\Omega_1} \cfrac{dN_2}{dx} AE \cfrac{dN_1}{dx}~dx + \int_{\Omega_2} \cfrac{dN_2}{dx} AE \cfrac{dN_1}{dx}~dx + \int_{\Omega_3} \cfrac{dN_2}{dx} AE \cfrac{dN_1}{dx}~dx \\ K_{13} & = \int_{\Omega_1} \cfrac{dN_3}{dx} AE \cfrac{dN_1}{dx}~dx + \int_{\Omega_2} \cfrac{dN_3}{dx} AE \cfrac{dN_1}{dx}~dx + \int_{\Omega_3} \cfrac{dN_3}{dx} AE \cfrac{dN_1}{dx}~dx \\ K_{14} & = \int_{\Omega_1} \cfrac{dN_4}{dx} AE \cfrac{dN_1}{dx}~dx + \int_{\Omega_2} \cfrac{dN_4}{dx} AE \cfrac{dN_1}{dx}~dx + \int_{\Omega_3} \cfrac{dN_4}{dx} AE \cfrac{dN_1}{dx}~dx \\ K_{22} & = \int_{\Omega_1} \cfrac{dN_2}{dx} AE \cfrac{dN_2}{dx}~dx + \int_{\Omega_2} \cfrac{dN_2}{dx} AE \cfrac{dN_2}{dx}~dx + \int_{\Omega_3} \cfrac{dN_2}{dx} AE \cfrac{dN_2}{dx}~dx \\ K_{23} & = \int_{\Omega_1} \cfrac{dN_3}{dx} AE \cfrac{dN_2}{dx}~dx + \int_{\Omega_2} \cfrac{dN_3}{dx} AE \cfrac{dN_2}{dx}~dx + \int_{\Omega_3} \cfrac{dN_3}{dx} AE \cfrac{dN_2}{dx}~dx \\ K_{24} & = \int_{\Omega_1} \cfrac{dN_4}{dx} AE \cfrac{dN_2}{dx}~dx + \int_{\Omega_2} \cfrac{dN_4}{dx} AE \cfrac{dN_2}{dx}~dx + \int_{\Omega_3} \cfrac{dN_4}{dx} AE \cfrac{dN_2}{dx}~dx \\ K_{33} & = \int_{\Omega_1} \cfrac{dN_3}{dx} AE \cfrac{dN_3}{dx}~dx + \int_{\Omega_2} \cfrac{dN_3}{dx} AE \cfrac{dN_3}{dx}~dx + \int_{\Omega_3} \cfrac{dN_3}{dx} AE \cfrac{dN_3}{dx}~dx \\ K_{34} & = \int_{\Omega_1} \cfrac{dN_4}{dx} AE \cfrac{dN_3}{dx}~dx + \int_{\Omega_2} \cfrac{dN_4}{dx} AE \cfrac{dN_3}{dx}~dx + \int_{\Omega_3} \cfrac{dN_4}{dx} AE \cfrac{dN_3}{dx}~dx \\ K_{44} & = \int_{\Omega_1} \cfrac{dN_4}{dx} AE \cfrac{dN_4}{dx}~dx + \int_{\Omega_2} \cfrac{dN_4}{dx} AE \cfrac{dN_4}{dx}~dx + \int_{\Omega_3} \cfrac{dN_4}{dx} AE \cfrac{dN_4}{dx}~dx ~. \end{align}$$ The matrix $$\mathbf{K}$$ is symmetric, so we don't need to explicitly compute the other terms.

From Figure 6, we see that $$N_1$$ is zero in elements 3 and 4, $$N_2$$ is zero in element 4, $$N_3$$ is zero in element 1, and $$N_4$$ is zero in elements 1 and 2. The same holds for $$dN_i/dx$$.

Therefore, the coefficients of the $$\mathbf{K}$$ matrix become
 * $$\begin{align}

K_{11} & = \int_{\Omega_1} \cfrac{dN_1}{dx} AE \cfrac{dN_1}{dx}~dx ~;~ K_{12} = \int_{\Omega_1} \cfrac{dN_2}{dx} AE \cfrac{dN_1}{dx}~dx ~;~ K_{13} = 0 ~;~ K_{14} = 0 \\ K_{22} & = \int_{\Omega_1} \cfrac{dN_2}{dx} AE \cfrac{dN_2}{dx}~dx + \int_{\Omega_2} \cfrac{dN_2}{dx} AE \cfrac{dN_2}{dx}~dx ~;~ K_{23} = \int_{\Omega_2} \cfrac{dN_3}{dx} AE \cfrac{dN_2}{dx}~dx  ~;~ K_{24} = 0 \\ K_{33} & = \int_{\Omega_2} \cfrac{dN_3}{dx} AE \cfrac{dN_3}{dx}~dx + \int_{\Omega_3} \cfrac{dN_3}{dx} AE \cfrac{dN_3}{dx}~dx ~;~ K_{34} = \int_{\Omega_3} \cfrac{dN_4}{dx} AE \cfrac{dN_3}{dx}~dx \\ K_{44} & = \int_{\Omega_3} \cfrac{dN_4}{dx} AE \cfrac{dN_4}{dx}~dx ~. \end{align}$$

We can simplify our calculation further by letting $$N^e_k$$ be the shape functions over an element $$e$$. For example, the shape functions over element $$2$$ are $$N^2_1$$ and $$N^2_2$$ where the local nodes $$1$$ and $$2$$ correspond to global nodes $$2$$ and $$3$$, respectively. Then we can write,
 * $$\begin{align}

K_{11} & = \int_{\Omega_1} \cfrac{dN^1_1}{dx} AE \cfrac{dN^1_1}{dx}~dx ~;~ K_{12} = \int_{\Omega_1} \cfrac{dN^1_2}{dx} AE \cfrac{dN^1_1}{dx}~dx ~;~ K_{13} = 0 ~;~ K_{14} = 0 \\ K_{22} & = \int_{\Omega_1} \cfrac{dN^1_2}{dx} AE \cfrac{dN^1_2}{dx}~dx + \int_{\Omega_2} \cfrac{dN^2_1}{dx} AE \cfrac{dN^2_1}{dx}~dx ~;~ K_{23} = \int_{\Omega_2} \cfrac{dN^2_2}{dx} AE \cfrac{dN^2_1}{dx}~dx  ~;~ K_{24} = 0 \\ K_{33} & = \int_{\Omega_2} \cfrac{dN^2_2}{dx} AE \cfrac{dN^2_2}{dx}~dx + \int_{\Omega_3} \cfrac{dN^3_1}{dx} AE \cfrac{dN^3_1}{dx}~dx ~;~ K_{34} = \int_{\Omega_3} \cfrac{dN^3_2}{dx} AE \cfrac{dN^3_1}{dx}~dx \\ K_{44} & = \int_{\Omega_3} \cfrac{dN^3_2}{dx} AE \cfrac{dN^3_2}{dx}~dx ~. \end{align}$$ Let $$K^e_{kl}$$ be the part of the value of $$K_{ij}$$ that is contributed by element $$e$$. The indices $$kl$$ are local and the indices $$ij$$ are global. Then,
 * $$\begin{align}

K_{11} & = K^1_{11} ~;~ K_{12} =  K^1_{12} ~;~ K_{13} = 0 ~;~ K_{14} = 0 \\ K_{22} & = K^1_{22} + K^2_{11} ~; K_{23} = K^2_{12} ~; K_{24} = 0 \\ K_{33} & = K^2_{22} + K^3_{11} ~;~ K_{34} = K^3_{12} \\ K_{44} & = K^3_{22} \\ \end{align}$$ We can therefore see that if we compute the stiffness matrices over each element and assemble them in an appropriate manner, we can get the global stiffness matrix $$\mathbf{K}$$.

Stiffness matrix for two-noded elements
For our problem, if we consider an element $$e$$ with two nodes, the local hat shape functions have the form

N^e_1(\mathbf{x}) = \cfrac{\mathbf{x}_2 - \mathbf{x}}{h_e} ~; N^e_2(\mathbf{x}) = \cfrac{\mathbf{x} - \mathbf{x}_1}{h_e} $$ where $$h_e$$ is the length of the element. Then, the components of the  element stiffness matrix are
 * $$\begin{align}

K^e_{11} & = \int_{x_1}^{x_2} \cfrac{dN^e_1}{dx} AE \cfrac{dN^e_1}{dx}~dx = \int_{x_1}^{x_2} \left(-\cfrac{1}{h_e}\right) AE                                  \left(-\cfrac{1}{h_e}\right)~dx = \cfrac{AE}{h_e} \\ K^e_{12} & = \int_{x_1}^{x_2} \cfrac{dN^e_2}{dx} AE \cfrac{dN^e_1}{dx}~dx = \int_{x_1}^{x_2} \left(\cfrac{1}{h_e}\right) AE                                  \left(-\cfrac{1}{h_e}\right)~dx = -\cfrac{AE}{h_e} \\ K^e_{22} & = \int_{x_1}^{x_2} \cfrac{dN^e_2}{dx} AE \cfrac{dN^e_2}{dx}~dx = \int_{x_1}^{x_2} \left(\cfrac{1}{h_e}\right) AE                                  \left(\cfrac{1}{h_e}\right)~dx = \cfrac{AE}{h_e} \end{align}$$ In matrix form,

{   \mathbf{K}^e = \cfrac{AE}{h_e} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} } $$ The components of the  global stiffness matrix are
 * $$\begin{align}

K_{11} & = \cfrac{AE}{h_1} ~; K_{12} =  -\cfrac{AE}{h_1} ~; K_{13} = 0 ~; K_{14} = 0 \\ K_{22} & = \cfrac{AE}{h_1} + \cfrac{AE}{h_2} ~; K_{23} = -\cfrac{AE}{h_2} ~; K_{24} = 0 \\ K_{33} & = \cfrac{AE}{h_2} + \cfrac{AE}{h_3} ~; K_{34} = -\cfrac{AE}{h_3} \\ K_{44} & = \cfrac{AE}{h_3} \\ \end{align}$$ In matrix form,

{   \mathbf{K} = AE \begin{bmatrix} \cfrac{1}{h_1} & -\cfrac{1}{h_1} & 0 & 0 \\ & \cfrac{1}{h_1} + \cfrac{1}{h_2} & -\cfrac{1}{h_2} & 0 \\ & & \cfrac{1}{h_2} + \cfrac{1}{h_3} & -\cfrac{1}{h_3} \\ \text{Symm.} & &  & \cfrac{1}{h_3}\\ \end{bmatrix} } $$

Load vector
Similarly, for the  load vector $$\mathbf{f}$$, we have
 * $$\begin{align}

f_j & = \int_0^L N_j\mathbf{q}~dx + \left. N_j\boldsymbol{R}\right|_{x=L} \\ & = \int_{\Omega_1} N_j\mathbf{q}~dx + \int_{\Omega_2} N_j\mathbf{q}~dx + \int_{\Omega_3} N_j\mathbf{q}~dx + \left. N_j\boldsymbol{R}\right|_{x=L} \end{align}$$ The components of the load vector are
 * $$\begin{align}

f_1 & = \int_{\Omega_1} N_1\mathbf{q}~dx + \int_{\Omega_2} N_1\mathbf{q}~dx + \int_{\Omega_3} N_1\mathbf{q}~dx + \left. N_1\boldsymbol{R}\right|_{x=L} \\ f_2 & = \int_{\Omega_1} N_2\mathbf{q}~dx + \int_{\Omega_2} N_2\mathbf{q}~dx + \int_{\Omega_3} N_2\mathbf{q}~dx + \left. N_2\boldsymbol{R}\right|_{x=L} \\ f_3 & = \int_{\Omega_1} N_3\mathbf{q}~dx + \int_{\Omega_2} N_3\mathbf{q}~dx + \int_{\Omega_3} N_3\mathbf{q}~dx + \left. N_3\boldsymbol{R}\right|_{x=L} \\ f_4 & = \int_{\Omega_1} N_4\mathbf{q}~dx + \int_{\Omega_2} N_4\mathbf{q}~dx + \int_{\Omega_3} N_4\mathbf{q}~dx + \left. N_4\boldsymbol{R}\right|_{x=L} \end{align}$$ Once again, since $$N_1$$ is zero in elements 2 and 3, $$N_2$$ is zero in element 3, $$N_3$$ is zero in element 1, and $$N_4$$ is zero in elements 1 and 2, we have
 * $$\begin{align}

f_1 & = \int_{\Omega_1} N_1\mathbf{q}~dx + \left. N_1\boldsymbol{R}\right|_{x=L} \\ f_2 & = \int_{\Omega_1} N_2\mathbf{q}~dx + \int_{\Omega_2} N_2\mathbf{q}~dx + \left. N_2\boldsymbol{R}\right|_{x=L} \\ f_3 & = \int_{\Omega_2} N_3\mathbf{q}~dx + \int_{\Omega_3} N_3\mathbf{q}~dx + \left. N_3\boldsymbol{R}\right|_{x=L} \\ f_4 & = \int_{\Omega_3} N_4\mathbf{q}~dx + \left. N_4\boldsymbol{R}\right|_{x=L} \end{align}$$ Now, the boundary $$\mathbf{x} = L$$ is at node 4 which is attached to element 3. The only non-zero shape function at this node is $$N_4$$. Therefore, we have
 * $$\begin{align}

f_1 & = \int_{\Omega_1} N_1\mathbf{q}~dx \\ f_2 & = \int_{\Omega_1} N_2\mathbf{q}~dx + \int_{\Omega_2} N_2\mathbf{q}~dx \\ f_3 & = \int_{\Omega_2} N_3\mathbf{q}~dx + \int_{\Omega_3} N_3\mathbf{q}~dx \\ f_4 & = \int_{\Omega_3} N_4\mathbf{q}~dx + \left. N_4\boldsymbol{R}\right|_{x=L} \end{align}$$ In terms of element shape functions, the above equations can be written as
 * $$\begin{align}

f_1 & = \int_{\Omega_1} N^1_1\mathbf{q}~dx = f^1_1\\ f_2 & = \int_{\Omega_1} N^1_2\mathbf{q}~dx + \int_{\Omega_2} N^2_1\mathbf{q}~dx = f^1_2 + f^2_1\\ f_3 & = \int_{\Omega_2} N^2_2\mathbf{q}~dx + \int_{\Omega_3} N^3_1\mathbf{q}~dx = f^2_2 + f^3_1\\ f_4 & = \int_{\Omega_3} N^3_2\mathbf{q}~dx + \left. N^3_2\boldsymbol{R}\right|_{x=L} = f^3_2 + \boldsymbol{R} \end{align}$$ The above shows that the global load vector can also be assembled from the element load vectors if we use finite element shape functions.

Load vector for two-noded elements
Using the linear shape functions discussed earlier and replacing $$\mathbf{q}$$ with $$a\mathbf{x}$$, the components of the element load vector $$\mathbf{f}^e$$ are
 * $$\begin{align}

f_1^e & = \int_{x_1}^{x_2} N^e_1 a\mathbf{x}~dx = \int_{x_1}^{x_2} \left(\cfrac{x_2 - x}{h_e}\right) a\mathbf{x}~dx = \cfrac{a}{h_e}\left(\cfrac{x_2(x_2^2-x_1^2)}{2}                                   - \cfrac{x_2^3-x_1^3}{3}\right) \\ f_2^e & = \int_{x_1}^{x_2} N^e_2 a\mathbf{x}~dx = \int_{x_1}^{x_2} \left(\cfrac{x - x_1}{h_e}\right) a\mathbf{x}~dx = -\cfrac{a}{h_e}\left(\cfrac{x_1(x_2^2-x_1^2)}{2}                                   - \cfrac{x_2^3-x_1^3}{3}\right) \end{align}$$ In matrix form, the  element load vector is written

{   \mathbf{f}^e = \cfrac{a}{h_e} \begin{bmatrix} \cfrac{x_2(x_2^2-x_1^2)}{2} - \cfrac{x_2^3-x_1^3}{3} \\ \cfrac{x_2^3-x_1^3}{3} - \cfrac{x_1(x_2^2-x_1^2)}{2} \end{bmatrix} } $$ Therefore, the components of the  global load vector are
 * $$\begin{align}

f_1 & = \cfrac{a}{h_1}\left(\cfrac{x_2(x_2^2-x_1^2)}{2} -                                \cfrac{x_2^3-x_1^3}{3}\right) \\ f_2 & = \cfrac{a}{h_1}\left(\cfrac{x_2^3-x_1^3}{3} -                                \cfrac{x_1(x_2^2-x_1^2)}{2}\right) + \cfrac{a}{h_2}\left(\cfrac{x_3(x_3^2-x_2^2)}{2} -                                \cfrac{x_3^3-x_2^3}{3}\right) \\ f_3 & = \cfrac{a}{h_2}\left(\cfrac{x_3^3-x_2^3}{3} -                                \cfrac{x_2(x_3^2-x_2^2)}{2}\right) + \cfrac{a}{h_3}\left(\cfrac{x_4(x_4^2-x_3^2)}{2} -                                \cfrac{x_4^3-x_3^3}{3}\right) \\ f_4 & = \cfrac{a}{h_3}\left(\cfrac{x_4^3-x_3^3}{3} -                                \cfrac{x_3(x_4^2-x_3^2)}{2}\right)+ \boldsymbol{R} \end{align}$$

Displacement trial function
Recall that we assumed that the displacement can be written as

\mathbf{u}_h(\mathbf{x}) = a_1\varphi_1(\mathbf{x}) + a_2\varphi_2(\mathbf{x}) + \dots + a_n\varphi_n(\mathbf{x}) = \sum_{i=1}^n a_i \varphi_i(\mathbf{x}) ~. $$ If we use finite element shape functions, we can write the above as

\mathbf{u}_h(\mathbf{x}) = a_1~N_1(\mathbf{x}) + a_2~N_2(\mathbf{x}) + \dots + a_n~N_n(\mathbf{x}) = \sum_{i=1}^n a_i~N_i(\mathbf{x}) $$ where $$n$$ is the total number of nodes in the domain. Also, recall that the value of $$N_i$$ is 1 at node $$i$$ and zero elsewhere. Therefore, we have
 * $$\begin{align}

u_1 & := \mathbf{u}_h(\mathbf{x}_1) = a_1~N_1(\mathbf{x}_1) + a_2~N_2(\mathbf{x}_1) + \dots + a_n~N_n(\mathbf{x}_1) = a_1 \\ u_2 & := \mathbf{u}_h(\mathbf{x}_2) = a_1~N_1(\mathbf{x}_2) + a_2~N_2(\mathbf{x}_2) + \dots + a_n~N_n(\mathbf{x}_2) = a_2 \\ \vdots \\ u_n & := \mathbf{u}_h(\mathbf{x}_n) = a_1~N_1(\mathbf{x}_n) + a_2~N_2(\mathbf{x}_n) + \dots + a_n~N_n(\mathbf{x}_n) = a_n \\ \end{align}$$ Therefore, the  trial function can be written as

\mathbf{u}_h(\mathbf{x}) = u_1~N_1(\mathbf{x}) + u_2~N_2(\mathbf{x}) + \dots + u_n~N_n(\mathbf{x}) = \sum_{i=1}^n u_i~N_i(\mathbf{x}) $$ where $$u_i$$ are the  nodal displacements.

Finite element system of equations
If all the elements are assumed to be of the same length $$h$$, the finite element system of equations ($$\mathbf{K}\mathbf{a} = \mathbf{f}$$) can then be written as

\cfrac{AE}{h} \begin{bmatrix} 1 & -1 & 0 & 0 \\     -1 & 2 & -1 & 0 \\      0 & -1 & 2 & -1 \\      0 & 0 & -1 & 1     \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \\ u_3 \\ u_4 \end{bmatrix} = \cfrac{a}{h} \begin{bmatrix} \left(\cfrac{x_2(x_2^2-x_1^2)}{2} - \cfrac{x_2^3-x_1^3}{3}\right) \\ \left(\cfrac{x_2^3-x_1^3}{3} - \cfrac{x_1(x_2^2-x_1^2)}{2}\right) + \left(\cfrac{x_3(x_3^2-x_2^2)}{2} - \cfrac{x_3^3-x_2^3}{3}\right) \\ \left(\cfrac{x_3^3-x_2^3}{3} - \cfrac{x_2(x_3^2-x_2^2)}{2}\right) + \left(\cfrac{x_4(x_4^2-x_3^2)}{2} - \cfrac{x_4^3-x_3^3}{3}\right) \\ \left(\cfrac{x_4^3-x_3^3}{3} - \cfrac{x_3(x_4^2-x_3^2)}{2}\right) + \boldsymbol{R}\cfrac{h}{a} \end{bmatrix} $$

Essential boundary conditions
To solve this system of equations we have to apply the  essential boundary condition $$\mathbf{u} = 0$$ at $$\mathbf{x} = 0$$. This is equivalent to setting $$u_1 = 0$$. The reduced system of equations is

\cfrac{AE}{h} \begin{bmatrix} 2 & -1 & 0 \\     -1 & 2 & -1 \\      0 & -1 & 1     \end{bmatrix} \begin{bmatrix} u_2 \\ u_3 \\ u_4 \end{bmatrix} = \cfrac{a}{h} \begin{bmatrix} \left(\cfrac{x_2^3-x_1^3}{3} - \cfrac{x_1(x_2^2-x_1^2)}{2}\right) + \left(\cfrac{x_3(x_3^2-x_2^2)}{2} - \cfrac{x_3^3-x_2^3}{3}\right) \\ \left(\cfrac{x_3^3-x_2^3}{3} - \cfrac{x_2(x_3^2-x_2^2)}{2}\right) + \left(\cfrac{x_4(x_4^2-x_3^2)}{2} - \cfrac{x_4^3-x_3^3}{3}\right) \\ \left(\cfrac{x_4^3-x_3^3}{3} - \cfrac{x_3(x_4^2-x_3^2)}{2}\right) + \boldsymbol{R}\cfrac{h}{a} \end{bmatrix} $$ This system of equations can be solved for $$u_2$$, $$u_3$$, and $$u_4$$. Let us do that.

Assume that $$A$$, $$E$$, $$L$$, $$a$$, and $$R$$ are all equal to 1. Then $$x_1 = 0$$, $$x_2 = 1/3$$, $$x_3 = 2/3$$, $$x_4 = 1$$, and $$h = 1/3$$. The system of equations becomes

\begin{bmatrix} 2 & -1 & 0 \\      -1 &  2 & -1 \\       0 & -1 &  1    \end{bmatrix} \begin{bmatrix} u_2 \\ u_3 \\ u_4 \end{bmatrix} = \begin{bmatrix} 0.037 \\     0.074 \\      0.383     \end{bmatrix} \qquad \implies \qquad \begin{bmatrix} u_1 \\ u_2 \\ u_3 \\ u_4 \end{bmatrix} = \begin{bmatrix} 0.0 \\ 0.494 \\ 0.951 \\ 1.333 \end{bmatrix} $$

Computing element strains and stresses
From the above, it is clear that the displacement field within an element $$e$$ is given by

\mathbf{u}^e = u^e_1 N^e_1(\mathbf{x}) + u^e_2 N^e_2(\mathbf{x}) ~. $$ Therefore, the strain within an element is

\boldsymbol{\varepsilon}^e = \frac{\partial \mathbf{u}^e}{\partial x} = u^e_1 \frac{\partial N^e_1}{\partial x} + u^e_2 \frac{\partial N^e_2}{\partial x} ~. $$ In matrix notation,

\boldsymbol{\varepsilon}^e = \mathbf{B}^e \mathbf{u}^e = \left[ \frac{\partial N^e_1}{\partial x} \frac{\partial N^e_2}{\partial x}\right] \begin{bmatrix} u^e_1 \\ u^e_2 \end{bmatrix} = \left[ -\cfrac{1}{h} \cfrac{1}{h} \right] \begin{bmatrix} u^e_1 \\ u^e_2 \end{bmatrix} $$ The stress in the element is given by

\boldsymbol{\sigma}^e = E\boldsymbol{\varepsilon}^e ~. $$ For our discretization, the element stresses are
 * $$\begin{align}

\boldsymbol{\sigma}^1 & = 1.48 \\ \boldsymbol{\sigma}^2 & = 1.37 \\ \boldsymbol{\sigma}^3 & = 1.15 \end{align}$$ A plot of this solution is shown in Figure 7.

Matlab code
The finite element code (Matlab) used to compute this solution is given below.


text: File:DistAxialMesh.png

title: File:DistAxialMesh.png


text: File:AxialBarFemDisp.png

title: File:AxialBarFemDisp.png


text: File:AxialBarFemSig.png

title: File:AxialBarFemSig.png


text:
User talk:JLCL

Hello JLCL, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --HappyCamper 02:13, 13 August 2007 (UTC)

WIkipedia and Wikiversity
Well, for me, I like to think of Wikiversity as a place where we are trying to develop online communities around online learning resources. I also think of Wikiversity as sort of like a collage. Here, we can potentially have different pages describing the same subject, whereas on Wikipedia, the aim is to have one representative (encyclopedic) article cover all the perspectives. Also, there are a lot more experimental pages on Wikiversity. Wikiversity is a much younger wiki compared to Wikipedia, so feel free to experiment with editing and organization of pages. There isn't a requirement that pages on Wikiversity require more information and details compared with Wikipedia, but editors here have tended to edit in this manner. HTH --HappyCamper 03:15, 15 August 2007 (UTC)

title: User talk:JLCL


text:
User talk:Saumilvadodaria

Hello Saumilvadodaria, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --HappyCamper 02:13, 13 August 2007 (UTC)

title: User talk:Saumilvadodaria


text:
File:Greenscreentest.ogg

Summary
I made this animation for How to use wiki technology as a free learner.

title: File:Greenscreentest.ogg


text:
File:Greenscreenexample.jpg

Summary
I made this image for How to use wiki technology as a free learner.

title: File:Greenscreenexample.jpg


text: Category:Computer security

title: Category:Computer security


text: Category:Organizations

title: Category:Organizations


text:
Interior design

This area is devoted to learning, teaching, and research related to interior design.

Discussion questions

 * What renowned interior designers have influenced you the most? How so?

Learning resources

 * /Words To Know/
 * /Selecting Furniture and Decorations/
 * Feng Shui
 * Green design

Wikipedia

 * Interior design
 * American Society of Interior Designers
 * Japanese interior design
 * Interior design regulation in the United States

title: Interior design


text:
User:RM77

I am bureaucrat on French Wikiversity. Please see my userpage for further informations.

Some pages :
 * Welcome templates

fr:User:RM77 it:User:RM77 de:User:RM77 es:User:RM77

title: User:RM77


text:
Talk:Filmmaking Basics/3D Storyboard/Introduction to FrameForge

On the FrameForge download page it says "20 runs max". What is a "run"? --JWSchmidt 14:18, 13 August 2007 (UTC)

You are allowed to open the program 20 times. The secret is to keep the program running in the background when you don't need it. As long as you do not quit the program, it only counts as a single run of the program. There is no other limits so you can open as many projects as you want. Most other programs are limited by the number of projects you create. With FrameForge 3D Studio demo version, you can close a project and then re-open it later and continue and it still counts as only a single run of the program. Robert Elliott 03:45, 14 August 2007 (UTC)


text:
User:Koala~enwikiversity

(8 December 1988) Hernán Manríquez. Born and living in Santiago, Chile. First year law student in the Pontificia Universidad Católica de Chile.

(Santiago de Chile, 8 de diciembre de 1988) Hernán Manríquez. También conocido como koala por los amigos.

Estudiante de Primer Año de Derecho en la Pontificia Universidad Católica de Chile.

title: User:Koala~enwikiversity


text:
Talk:Pseudoscience

The title is incorrectly spelled. Please correct from "Psuedoscience" to "Pseudoscience".
 * Done. Thanks for the note. --HappyCamper 19:59, 13 August 2007 (UTC)

can of worms
Long standing revert wars and other editorial conflicts arose on Wikipedia over this topic. Studying the topic of pseudoscience (i.e, non-science masquerading as science) is appropriate on Wikiversity, but this topic is often dogmatically approached, by "debunkers," whose focus is on discrediting what they disagree with. See Pseudoskepticism and our resource Pseudoskepticism for some balance.

"Pseudoscience" is sometimes loosely used to refer to any non scientific belief or practice, including religion or quasi-religious belief.

The tag gets applied to, for example, the Paranormal. Consider CSICOP, the "Committee for the Scientific Investigation of Claims of the Paranormal." At least some of the founders of CSICP were true skeptics, who did not forget to be skeptical of their own beliefs as well, hence "scientific investigation," which is, by definition, open-minded. It proceeds through testing of claims, and never settles on a particular test result as utterly and finally conclusive. Science is not about forming belief, but about forming predictive theory and testing it. When a test confirms a theory, the theory progresses toward acceptance or further acceptance, normally, and when it disconfirms a theory, the theory regresses, that's all. It is always possible that the theory wasn't exactly wrong, but incomplete, the test encountered a not-understood exception, etc.

This is all well-known to genuine scientists and sociologists of science, but not necessary well-understood by contemptuous debunkers.

Many "paranormal" claims can be scientifically tested. Some might not be. That does not make the paranormal itself pseudoscientific, because the paranormal is an asserted class of exceptional phenomena, and is not a specific theory or body of belief. A specific theory of the paranormal may be pseudoscientific, taking on the trappings of science, but the judgment of "pseudoscience" is often simply subjective polemic.

We will study pseudoscience as a topic here. We may study the application of that label to various fields and social practices, including "debunking," which often makes pseudoscientific claims.

Carl Sagan wrote Carl Sagan,
 * It seems to me what is called for is an exquisite balance between two conflicting needs: the most skeptical scrutiny of all hypotheses that are served up to us and at the same time a great openness to new ideas … If you are only skeptical, then no new ideas make it through to you … On the other hand, if you are open to the point of gullibility and have not an ounce of skeptical sense in you, then you cannot distinguish the useful ideas from the worthless ones. --"The Burden of Skepticism" in Skeptical Inquirer Vol. 12, Issue 1 (Fall 1987)

Richard Dawkins is reported as having written:
 * There's this thing called being so open-minded your brains drop out.

He was quoting someone else, obviously. I have seen the comment attributed to Carl Sagan, but have also seen a story that someone said it to Sagan when he made some friendly, open-minded approach or comment on those investigating UFOs. I.e., they were warning Sagan. I.e., there are obviously ideas that are so dangerous that we should not be open-minded about them, lest our minds rot and our brains drop out.

Interesting idea, eh? There is such an idea, and it's very simple: "You are wrong, I'm right." It definitely rots the brain, causing it to become highly dysfunctional, serving only the cause of uncovering "the proof" that I'm right and you are wrong. All other evidence is discarded as misleading or irrelevant.

I found this: which provides an early reference from 1937. --Abd (discuss • contribs) 20:43, 9 April 2014 (UTC)

title: Talk:Pseudoscience


text:
User:Luai lashire

My Goals For Wikiversity
Now that I've been around Wikiversity for a while, I've got a clearer sense of where we are now and what changes we might want to implement. The biggest thing, to me, is focusing on getting away from the blah-blah lecture-y type course we are used to. Projects like the Bloom Clock make great use of the wiki technology and break away from the traditional style of learning. Other good things are begining to emerge, but need to be used more; reading groups, video and audio, and lots and lots of discussion should be a part of almost every learning resource. More inter-resource linking needs to be done to get people moving from page to page on Wikiversity. I want to find ways for us to make learning resources more interactive and engaging and more student-led than they currently are. Discussion is only one way of doing this. I am currently toying with some ideas I want to implement on The Neurodiversity Movement learning resource that would do this. If they are effective, I hope they can be adapted to suit many other learning resources. These ideas include having the students do research and write part of the course themselves, hunt down and discuss related news articles or other resources on the web, and repeatedly sending the students to other websites to read the original sources for the material in the course. I'm very busy during the school year, but when I have the time, I will be working more on my Neurodiversity Movement resource, adding more content to the Writing Portal, and hopefully starting a collaborative writing community that will try to write a novel (or possibly shorter piece) collaboratively using the wiki technology most effectively.

A Note on Miscommunication
I do have Asperger's Syndrome, and communication is something that is sometimes difficult for me. While I have a very good grasp of the technicalities of writing, I don't always understand how to phrase something right, to have the best impact on the reader. If I offend anyone in any way, I can assure you it's unintentional; I try very hard to be completely open and discuss things without anger or bias, and I do not ever intentionally write inflamatory statements. If something I type seems blunt, harsh, or insulting, please, PLEASE ask me about it on my talk page; that way we can clear up the miscommunication. I also sometimes am not sure of the rules or the way things are typically done, so if I do something "wrong" or you feel I should have contacted you about some change I make- or if I DO contact you about something that I didn't need to- please bring it up on my talk page if you feel it needs to be discussed. Always assume good faith!

My websites

 * DeviantArt
 * My DeviantArt club for Autistic Pride art
 * FictionPress
 * Polyvore

My involvement in Wikiversity
I am a participant in the Bloom Clock learning project.

I have taken an interest in the Quote of the Day learning project, and while I don't expect to be very deeply involved in it, I have been participating some in the discussion.

I am currently using most of my free time to read Ivan Ilich's Deschooling Society with the hope of somewhat revitalizing the current reading group for that book, where discussion has lagged.

I am also working on developing a learning resource, called The Neurodiversity Movement, about the civil rights movement of the same name. Anyone who wishes to help with this project is more than welcome, particularly those with more experience here at Wikiversity; the formatting of pages is something that gives me a bit of trouble.

You can see my responses to the Wikiversity user survey questions here.

Contact
You can contact me on my user talk page or, if for some reason that isn't appropriate, you can always reach me through my e-mail.

title: User:Luai lashire


text:
User talk:Luai lashire

Bloom Clock Category Group
Hi Luai. Nice to see so many entries on the bloom clock! I'm wondering what region the category should cover for your contributions... any ideas? There is a category for Southeastern PA now (covering from the southeast corner of the state to the foothills), but I don't know enough about the differences in the center of the state for bloom time (i.e., what counties would see the plants blooming within a week or so of plants blooming in Centre County in early spring and late fall). Would "south-central PA" be right? I'm guessing that north of you will be affected by elevations and possibly lake effects, but I really don't know.

Categories are used for enabling DPL lists, which in turn can be used to create the "keys" (like this). --SB_Johnny | PA! 13:12, 19 August 2007 (UTC)

 * Hi. I think it might be best to have a "central PA" catagory.  I suppose if you wanted to be really specific you could divide into smaller catagories, but considering that I am the only contributor from central PA right now, I don't think that's necesary at this point.
 * As far as differences in bloom time in the central counties goes, I don't know a lot about it; but I do know that in the centre region it varies a lot even quite locally. I live on the edge of State College; in State College proper, Sedum is already blooming, but the Sedum in my garden won't be blooming for several more weeks.  --Luai lashire 17:46, 19 August 2007 (UTC)

Various stuff...
Hi Luai. Just an update on the categories first: I made Category:BCP/CTPA/8 for plants you've logged this month, as well as a basic visual key (Bloom Clock/Flowers seen in August in Central Pennsylvania). Adding categories to the plant profile pages will allow the software to pick them up and add them to the keys on the next update (if I'm not in touch with you before then, September's logs should use Category:BCP/CTPA/9).

On another bloom clock topic, don't forget that while the Genus is capitalized, the species is not (which is why you got a redlink for Trifolium Pratense... I fixed it to read Trifolium pratense :-).

Finally, we've been hit by vandals several times over the weekend. If you don't mind, please vote on Nominations for checkuser/SB Johnny and Nominations for checkuser/JWSchmidt... checkuser lets us track down the IPs used for vandalism, which can help prevent further attacks. --SB_Johnny | PA! 11:51, 26 August 2007 (UTC)

 * Hi! The CTPA pages look good, except that some of the plants I've logged aren't on them.  Is this just because those plants don't yet have log pages? --Luai lashire 15:24, 26 August 2007 (UTC)

 * Yup. I'm working on making log page creation easier, but the template is rediculously complicated and I'm not the best person for explaining technical stuff that I just barely understand myself :P. I'll leave you a link later... just popping in for a sec right now. --SB_Johnny | PA! 16:38, 26 August 2007 (UTC)

 * I did one of the category additions for September... don't have much time today to do the others. this diff will show you how to alter the others. --SB_Johnny | talk 12:09, 2 September 2007 (UTC)

See Bloom Clock/Project Discussion for more stuff. --SB_Johnny | talk 10:23, 4 September 2007 (UTC)

For younger students only?
I'm woefully ignorant in a number of areas (a product of 4 years' formal education total) in which a course for 8-14 year-olds may be my level! Would I be excluded? Shir-El too 11:48, 4 September 2007 (UTC)
 * To what are you referring? Wikiversity in general?  The Neurodiveristy course?  --Luai lashire 12:28, 4 September 2007 (UTC)

Finding collaborators...
Don't feel bad... I've often felt the same way on Wikibooks (where I work on a garden book when I'm not racing around fixing the bloom clock), but on Wikibooks most editors are actually lone wolves (sole authors of one of more books), so I was able to get used to it with moral support there :).

I do have a friend with a talk-page spamming bot on Wikipedia: if you think it's a good idea, he might be willing to target w:Category:Wikipedians_on_the_autism_spectrum for you. --SB_Johnny | talk 23:36, 29 September 2007 (UTC)
 * Yes, please! I tried to drum up support at a couple of autism pride websites I go to, but no one there was interested in Wikiversity, so it failed.  This bot idea would probably have much better luck.  --Luai lashire 00:51, 1 October 2007 (UTC)
 * I think I'm the friend who Johnny is talking about. :-) I'd be happy to do this if you just typed up something so send to them. :-)  Be sure to include:
 * What Wikiversity is.
 * What exactly you want to do.
 * A few good links. :-)
 * If you like this idea/when you write it, please feel free to e-mail me.  Cbrown1023   talk   23:36, 1 October 2007 (UTC)
 * Thanks! I'll get right on that.  --Luai lashire 20:48, 2 October 2007 (UTC)

Fundamental Information is Irrevelent?
Can you explain how a link to fundamental information regarding Bipolar disorder is irrevelent to "neurodiversity" when you are using as a fundamental category? Do you intend to reformulate basic definitions? Mirwin 21:29, 1 October 2007 (UTC)
 * I'm sorry, I was actually just about to leave you a message explaining why I removed that link, but you beat me to the punch. It's not that the link was "irrelevant"- it's that it was not suited to the purpose of the course.  The overviews of each condition in the first section of the course, when they are completed, are meant to cover basic information of what the condition is, in medical terms, and impart an understanding of why that condition is being deemed a neurotype.  As such the information can't be described in an overly negative light, as it was in your resource, or it simply doesn't make sense in the context of the course.  I can't respond to you properly right now, I have to leave, but I'll respond in more detail when I get back.  --Luai lashire 21:49, 1 October 2007 (UTC)

more...
Someone's been working on Aspies Book... maybe find someone there?

BTW, any flowers this month? :P --SB_Johnny | talk 14:49, 5 October 2007 (UTC)
 * I did see a few, but I actually haven't been outside much since school started up again- I spend the day inside there, then come home and spend the evening on homework. It's exhausting.  I'm trying to find ways to free up some more time, and when I do, I'll have more flowers to log.  :)  --Luai lashire 21:11, 5 October 2007 (UTC)
 * Of course, now I can't walk. It figures!  Yesterday I hurt my foot very badly and I have to stay off it for a week or so....  So no flowers.  Sorry.  --Luai lashire 14:43, 6 October 2007 (UTC)

keymaking
Hi Luai. There's a few steps to making keys (I really need to make up some general instructions some time).

 * 1) Add categories to the BCP pages after you sign the logs. For example, since you signed BCP/Sedum spectabile/Logs, you would add 2 templates to BCP/Sedum spectabile, like this. The template bcpm/10 adds "October" to the regional data for Central PA, and bcp/ctpa/10 adds it to Category:BCP/CTPA/10 (as well as other categories later).
 * 2) Next you make a DPL page. I'll do that for you this month so you can see what I did.
 * 3) Finally, you copy the list from the DPL result (as plain text, copying from the resource page, not the edit page), and add colons and brackets which in turn make the BCP pages act like templates on the key.

The first step is the most important one (I'm happy to generate the DPLs and keys... it's become second nature, so to speak). You can find all the logs you've signed using Special:Contributions/Luai_lashire, or using Special:Whatlinkshere/User:Luai_lashire. --SB_Johnny | talk 11:52, 13 October 2007 (UTC)
 * OK! :)  I've started adding the catagories I need to the BCP pages.  --Luai lashire 01:39, 14 October 2007 (UTC)
 * Here's the keys: Bloom_Clock/Keys/Central_Pennsylvania. If you see any you don't know this month, they'll probably be on one of the SEPA lists for October: Bloom_Clock/Keys/Southeastern_Pennsylvania/October--SB_Johnny | talk 09:24, 14 October 2007 (UTC)
 * Thanks. Oh- I found this website a while ago; it makes an excellent guide for identifying flowers in my local area.  That's where I look if I can't find them in the SEPA keys, and it's only let me down once.  --Luai lashire 14:22, 14 October 2007 (UTC)

collaborating with other writers
"collaborating with other writers" <-- I like the idea that Wikiversity can specialize in "learn by doing" projects. I've started Science Fiction Challenge and Exploring science through fiction, and it seems like there could be similar projects for fantasy. Do you think such efforts should all be coordinated through the Writing Center? --JWS 14:05, 28 October 2007 (UTC)
 * I think that, in general, the more wikiversity's various centres, portals, courses, and groups coordinate and comunicate, the easier it will be for interested parties to find what they're looking for, so yes, it could be coordinated through the Writing Center. There are different forms this project could take:  There's collaboration to produce one single, finished piece, or collaboration to help others find inspiration to start or finish a piece, or just a critique group, ect.  There are different ways we could do a collaboration; in the past, the only collaborative work I've helped with was Winding lane, a story project utilizing LiveJournal that never got finished.  I'm particularly fond of this kind of format for colaboration, where each writer adds a new "chapter" or section of work, but we could do others as well.  What are your thoughts?  --Luai lashire 00:55, 30 October 2007 (UTC)
 * The only place I've done collaborative fiction writing is at the fiction Wikia. I'm willing to try just about any format.--JWSchmidt 01:10, 30 October 2007 (UTC)

Solar System grammar - audio message
--Juan 20:33, 3 December 2007 (UTC)

Regarding Bloom Clock Orphans
Luai, orphan pages may be found with the utility on the "Special Pages" menu. It is to the lower left under the search box about seven items down on the left column menu on the default Wikiversity view. After you click this there is a long list of special pages/utilities, IIRC they are alphabeted. Let me know if you have any trouble finding the utility from these directions. Merry Christmas and Happy New Year to you and yours! Mirwin 00:50, 26 December 2007 (UTC)

Hello :-)
Hi Luai, just dropping by and saying hello really. :-) I liked what I just read on your userpage - hadn't been on it for a while. I'm also trying to get around to reading Illich, and will put up notes on the page - I also had a look at the Neurodiversity page, but didn't feel able to just jump in, as you'd obviously like me to do. Perhaps I can help you figure out a way for the project to be more inviting? Cormaggio talk 10:24, 25 January 2008 (UTC)
 * Sure! That would be really helpful.  I had hoped that the questions and invitation to discuss would help draw people in, but obviously not.  I'm not always good at conveying feeling in my writing, is that part of the problem?  Or is it more an issue of the course's structure?  --Luai lashire 17:52, 25 January 2008 (UTC)

Luai, i've been a bit behind on the reading for Illich lately but at this point I think i'll take your lead and jump into chapter 6 on learning webs. I think one of the nice things about doing a reading group this way is how it can materialise over quite some time. Countrymike 20:00, 25 January 2008 (UTC)

 * On drawing people in to the Neurodiversity page, I think what's needed is a bit more clarity on what exactly people are being asked to discuss. It might be difficult for people to give opinions on autism if they have no real experience with it. Perhaps some illustrative 'case study' examples, and some guidance in starting a discussion might be helpful - perhaps with some prompter questions (though that's only one possibility here). I'll try to give some more specific feedback on its talk page - and don't be shy to remind me if I don't. :-) Cormaggio talk 12:31, 26 January 2008 (UTC)
 * Aaahhhh. Thank you, Cormaggio, I think I understand my problem now.  One of my greater faults is that I tend to omit information without realizing it- It's hard for me to re-trace my own steps of understanding and explain things in a way that is fully understandable to others.  It sounds like that's what I'm doing; the course, as it stands right now, makes complete sense to me and is already at a stage where I can discuss it, but then, I also have other information which the students don't yet, and now I have to figure out which parts need to be incorporated into the course to make it whole.  (sorry if this doesn't entirely make sense, I'm very tired right now!)  --Luai lashire 03:37, 27 January 2008 (UTC)
 * I understand what you're saying, but you don't need to be so hard on yourself. :-) That's what this wiki environment is best for - for people to point out things that we may ourselves have become blind to - and it happens to us all... Cormaggio talk 12:09, 27 January 2008 (UTC)

Just curious
Do you consider bipolarism to be a neurotype? --SB_Johnny | talk 23:29, 1 February 2008 (UTC)
 * Yes, I do. I don't have it myself, so I don't have personal experience of it, but I know there's fairly widespread support in the Neurodiversity community for it.  One of the Autism Hub bloggers, Kevin Leitch (who unfortunately no longer blogs, due to threats made towards his daughter), is bipolar and used to talk about it on and off on his blog, so I know that he views there being some very good traits associated with it.  He says that when he is in a mixed state, between manic and depressed, he is at his most creative, and that without this he would not be nearly as good as he is at his job.  He also acknowledges that bipolarism needs to be handled a certain way, otherwise you are risking putting yourself in danger.  He talks about it on this blog sometimes, although he doesn't update very much.  The posts there are worth looking through, though, if you're interested.  --Luai lashire 17:12, 3 February 2008 (UTC)
 * Interesting. More than anything else I'm acutely seasonally affective. I'm at my most creative from a few weeks after winter solstice to a week or so after summer solstice. Then I'm more expressive as the days get shorter. I think :). --SB_Johnny | talk 17:56, 3 February 2008 (UTC)

Wikimedian Demographics update
Hello Luai lashire. I'm writing because you requested a reminder about progress on Wikimedian Demographics. For the time being these will be about once per month.

Speaking of reminders, it turns out the template question isn't the most efficient way of coming up with a mailing list. If you want to recieve this message on another of your talk pages (or don't want them at all), please edit the list on Wikimedian Demographics/Reminders.

To update on progress so far, there have been somewhere around 50 respondents over the past month, with more trickling in slowly but steadily. There are 7 major surveys (or 12, depending on how you count them):
 * 1) The basic survey (about your participation in Wikimedia)
 * 2) The "geography" surveys (actually 6 of those, for each inhabited continent)
 * 3) The Operating System survey (still stubby)
 * 4) The Wikipedia survey
 * 5) The Commons survey
 * 6) The US presidential elections survey
 * 7) The sexuality and relationships survey

Only the geography surveys have had any major modifications since being originally posted (more countries were added to those).

Several new surveys have been discussed at Talk:Wikimedian Demographics, as well as other places within Wikiversity. A few that might be ready sometime in February include:
 * A survey about Women in Wikimedia
 * A survey about Wikibooks
 * A survey about Wikiversity
 * A survey about administrative issues (including cross-wiki topics such as CheckUser and "global blocking")
 * A survey about environmental issues
 * A survey about religion
 * A survey about international trade

If you have ideas for more surveys, please share them on Talk:Wikimedian Demographics. Thanks for your participation thus far, and we hope to see you soon! --SB_Johnny | talk 11:58, 5 February 2008 (UTC)

Email contact
Hi Luai,

To fend off spammers, you might want to use a link (Special:Emailuser/Luai_lashire) instead :). Of course that requires people to have an account, but it only takes a few minutes to make those anyway. --SB_Johnny | talk 15:38, 21 February 2008 (UTC)
 * OK. I thought it would probably be enough just to leave it broken up the way I did, but since there's a better option I'll just change it.  Thanks.  :)  --Luai lashire 04:16, 22 February 2008 (UTC)

Templates :)
Hey Luai! Good to see stuff blooming out in CTPA... don't forget to add the templates to the profile pages though :). I fixed and  this morning. They're easy to make if you need to in the future: just copy-paste from another month's template, and switch the month name and number (each appears 3 times on the template).

I've been enormously busy this spring, but I'm trying to set up the keys to be much easier to construct. The global keys are coming along quite well also: judging by what you logged in the past few days, you should be able to use them to identify unknown plants using either Bloom Clock/Keys/Global/Temperate/Late Winter or Bloom Clock/Keys/Global/Temperate/Early Spring. I'll set up a data comparison page for you in the next few days. --SB_Johnny | talk 10:38, 12 April 2008 (UTC)

Asparagus
Hi Luai. Me too. How do I add myself to the aspergian wikipedians? Cheers Janet
 * You can add on your user page: . Did you see also this page at meta (and the talk page) ? Erkan Yilmaz uses the Wikiversity:Chat (try) PS: Tag a learning project with completion status !! 04:49, 31 May 2008 (UTC)---Erkan Yilmaz uses the Wikiversity:Chat (try) PS: Tag a learning project with completion status !! 04:49, 31 May 2008 (UTC)

Thanks Erkan lucychili 00:06, 1 June 2008 (UTC)

Re: Perhaps not the right category?
Thank you for your communication. I wasn't altogether sure what the neurodiversity movement was so I thank you for informing me and I would be happy to learn more about it. I saw that one of your pages contained information on Autism so I categorized according to my own teaching on the subject. The problem is that there are a lot of medical conditions that overlap between neurology and psychiatry, of which autism is one. It was my understanding that it lies more in the study of psychiatry than in the study of neurology, but it should definitely be categorized as both. Your pages, however, seem to be based on a sociological movement and should maybe also be classfified as sociology (a science) as well. Please feel free to discuss this with me further so we can reach consensus. Donek (talk) - Go raibh mile maith agaibh 22:46, 20 July 2008 (UTC)

title: User talk:Luai lashire


text:
User:JohnManuel

es:Usuario:JohnManuel

title: User:JohnManuel


text:
User talk:JohnManuel

Hello JohnManuel, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --HappyCamper 20:05, 13 August 2007 (UTC)

title: User talk:JohnManuel


text: File:BoeBot Side.JPG

title: File:BoeBot Side.JPG


text: Introduction to Robotics/Visuals/BoeBot

title: Introduction to Robotics/Visuals/BoeBot


text:
Talk:Introduction to psychology/Questions one: Definition and History

"Expand"
In question 1, the "expand" links are hidden behind the text of the question. I don't know how to fix this. --Luai lashire 21:13, 13 August 2007 (UTC)

 * You can try the alternative (but slightly buggy) approach in Help:Quiz-Simple -- User:Banerjee

double content
see here please, Erkan Yilmaz ( Wikiversity:Chat, wiki blog ) 17:55, 16 January 2008 (UTC)


text:
User:Smaug

I am Dragon Smaug on wikipedia and Smaug on mediawiki.

title: User:Smaug


text:
User talk:Smaug

Feel free to talk to me.

title: User talk:Smaug


text:
Markerless Tracking

The Markerless Tracking project is doing original research in the field of computer based object recognition and tracking. The aim of this project is to archive a general and widely accepted technique to perform tracking and recognition of the real environment without using any special placed markers eg. fiducials.

In order to achieve this goal we will analyze the captured reality by comparing the data to a huge number of on-the-fly generated possible configurations. This approach is called Analysis-by-Synthesis (AbS).

Motivation
A big challenge in Computer Vision is the recognition and tracking of real objects through sensor-data streams. This is needed for example in Robotics and Augmented Reality to gather information about the surrounding environment. Standard Computer Vision techniques gives good results if the objects are very simple and the sensor data is not too biased. Generally spoken we have a huge amount of computer systems which works well on recognizing and tracking special fiducial markers which must be distributed in the environment. Unfortunately there are only a few special cases where markerless tracking and recognition could be archived with todays techniques.

About Tracking
''In its simplest form, tracking can be defined as the problem of estimating the trajectory of an object in the image plane as it moves around a scene. In other words, a tracker assigns consistent labels to the tracked objects in different frames of a video. Additionally, depending on the tracking domain, a tracker can also provide object-centric information, such as orientation, area, or shape of an object.''

Idea

The basic idea is very simple. Just

 * 1) capture the reality through sensor data (e.g. an camera image),
 * 2) generate all possible virtual sensor datasets with the computer (e.g. computer graphics rendering) and
 * 3) select the virtual dataset which is most similar with the captured one for getting the current configuration in reality (e.g. position and orientation of the camera).

This sounds not very high performance and maybe seems to be not accurate enough for special purposes. Also a potential implementation may be questionable at the beginning. Please read about the resarch questions and answers below before you abandon the whole idea.

Assumptions
As far as known this technique works under with three assumptions:

 * 1) the reality can be captured from the computer through sensors
 * 2) the computer can simulate possibly captured configurations of the reality
 * 3) captured and generated datasets can be compared in a meaningful way

Research questions and answers

 * How to increase the performance of the process?
 * Which assumptions will introduce errors and what is the overall accuracy?
 * Could this idea be realized with todays soft and hardware?
 * What is the best method to compare special kinds of data sets for selection?
 * Which research articles are helpfull and what kind of work is related to this project?
 * How can markless tracking be used in Digital Learning Environments

How to contribute
Like all research projects this one really needs collaboration between people. So first thing you could do is being bold while reading these wikipages. Go ahead and edit the content! Especially if you have questions or disagree with given statements. Don't forget to write your name into the list of contributors below.

There a also a lot of places which need to be filled up with more thoughts, comments and ideas. Look at the research questions above for a start.

Learning Tasks

 * (TrackingJS]) Explore the features of the Open Source Library TrackingJS and explain how markless tracking can be used in Digital Learning Environments
 * (Markerbased Tracking) Compare Markerless Tracking with Markerbased Tracking. Describe the differences and similarities. Analyze PROs and CONs of the approach for different applications (motion analysis, particle tracking, ...)
 * (GeoAR) Analyse the concept of geolocation based augmented reality and describe the role of markerless tracking in that context. Analyze the features and performance of mobile devices and define possible options of application for tracking in conjunction with geolocated information.

Contributors

 * Rodja Trappe

title: Markerless Tracking


text:
Talk:Markerless Tracking

Just started
Now i've done the first step. I wanted to start this research project here at wikiversity for long. In the next days i'll provide more details about how the aim could be archived, what questions arise with this and how other people could contribute. Trappe 22:22, 13 August 2007 (UTC)

title: Talk:Markerless Tracking


text:
User:Yulyan~enwikiversity

Yuyan's page

Hi. My name is Yulyan. I heard about Wikis and got really interested in learning how to use this tool. I'm a language teacher. I teach English in high school and college (business English). I'm willing to learning a lot from this page and share teaching experiences as well.

title: User:Yulyan~enwikiversity


text:
File:Astroboi Seduced By The Dark Side.pdf

Used with the WikiU Film School Lesson on how to format a movie script


text:
Nonlinear finite elements/Model problems

Model problems
Let us look at a couple of model problems.

Problem 1: First-order homogeneous ODE
The first problem involves a homogeneous first order differential equation and is stated as:
 * $$\text{(1)} \qquad

{     \begin{align} \text{Find}~ u(t) & ~\text{such that it satisfies} \\ & \frac{du}{dt} + u = 0, \qquad t > 0 \\ & u(0) = 1 \end{align} } $$ The exact solution is
 * $$\text{(2)} \qquad

{     u(t) = e^{-t}~. } $$

Problem 2: Second-order inhomogeneous ODE
The second problem involves an inhomogeneous second order differential equation and is stated as:
 * $$\text{(3)} \qquad

{    \begin{align} \text{Find}~ u(x) & ~\text{such that it satisfies} \\ & -\frac{d^2u}{dx^2} + u = x, \qquad x \in (0,1) \\ & u(0) = 0, \qquad u(1) = 0 \end{align} } $$ The exact solution is
 * $$\text{(4)} \qquad

{     u(x) = x - \frac{\sinh(x)}{\sinh(1)} ~. } $$

Remarks
Both model problems have exact solutions. However, in most situations, such solutions may be impossible to get because:
 * No solution exists since the data are not smooth enough.
 * Even if a solution exists, it cannot be found in closed form because of the complexity of the problem.

To overcome these problems, we can do the following :
 * Reformulate the problem so that it admits  weaker conditions on the solution and its derivatives. Thus the equation no longer needs to be satisfied at  all points in the domain.  These are called  weak or  variational formulations.
 * Break the domain up into smaller pieces and require the equation to be satisfied at only a few points within each piece.
 * Assume that the approximate solution has a known form.

Finite element methods use a weak (or variational) formulation of the original  strong problem.


text:
File:Storyboard Sample Refardeon.png

Used with the WikiU Film School Lesson on how to do thumbnail storyboarding

title: File:Storyboard Sample Refardeon.png


text:
File:Storyboard Refardeon.png

Used with the WikiU Film School Lesson on how to do thumbnail storyboarding

title: File:Storyboard Refardeon.png


text:
Nonlinear finite elements/Weighted residual methods

Weak Formulation : Weighted Average Methods
Weighted average methods are also often called "Rayleigh-Ritz Methods". The idea is to satisfy the differential equation in an  average sense by converting it into an integral equation. The differential equation is multiplied by a weighting function and then averaged over the domain.

If $$v(t)$$ is a weighting function then the  weak form of Equation (1) is
 * $$\text{(5)} \qquad

{   \int^1_0 \left(\frac{du}{dt} + u\right)~v(t)~dt = 0 ~;\qquad u(0) = 1 ~. } $$

The weighting function $$v(t)$$ can be  any function of the independent variables that is sufficiently well-behaved that the integrals make sense.

Recall that we are looking for an  approximate solution. Let us call this approximate solution $$u_h$$. If we plug the approximate solution into equation (5) we get
 * $$\text{(6)} \qquad

{   \int^1_0 \left(\frac{d}{dt}(u_h) + u_h\right)~v(t)~dt = \int^1_0 R(t)~v(t)~dt = 0 ~; \qquad u_h(0) = 1 ~. } $$

Since the solution is approximate, the original differential equation will not be satisfied exactly and we will be left with a  residual $$R(t)$$. Weighted average methods try to minimize the residual in a weighted average sense.

Finite element methods are a special type of weighted average method.

Examples of Weighted Average Methods
Let us assume the trial solution for problem (6) to be

u_h(t) = a_0 + a_1 t + a_2 t^2 + \dots + a_n t^n ~. $$ After applying the initial condition we get $$a_0 = 1$$, and the trial solution becomes

u_h(t) = 1 + a_1 t + a_2 t^2 + \dots + a_n t^n ~. $$ Let us simplify the trial solution further and consider only the first three terms, i.e.,
 * $$\text{(7)} \qquad

{   u_h(t) = 1 + a_1 t + a_2 t^2~. } $$ Plug in the trial solution (7) into (6). Then, the residual is
 * $$\text{(8)} \qquad

{   R(t) = 1 + a_1(1+t) + a_2(2t + t^2)~. } $$ If $$R(t) = 0$$, then the trial solution is equal to the exact solution. If $$R(t) \ne 0$$, we can try to make the residual as close to zero as possible. This can be done by choosing $$a_1$$ and $$a_2$$ such that $$R(t)$$ is a minimum.

Minimizing R(t): Collocation Method
In the collocation method, we minimize the residual by making it vanish at $$n$$ points $$t_1, t_2,\dots, t_n$$ within the domain.

For our problem, the domain of interest is $$0 \le t \le 1$$. Let us pick two points in this domain $$t_1$$ and $$t_2$$ such that $$0 \le t_1 < t_2 \le 1$$ (see Figure 1). In this example we choose $$t_1 = 1/3$$ and $$t_2 = 2/3$$.

The values of the residual (8) at $$t_1$$ and $$t_2$$ are

R(t_1) = 1 + \frac{4}{3} a_1 + \frac{7}{9} a_2 ~; \qquad R(t_2) = 1 + \frac{5}{3} a_1 + \frac{16}{9} a_2 ~. $$ If we now impose the condition that the residual vanishes at these two points and solve the resulting equations, we get $$a_1 = -27/29$$ and $$a_2 = 9/29$$. Therefore the approximate solution is

{     u_h(t) = 1 - \frac{27}{29} t + \frac{9}{29} t^2~. } $$ Figure 2 shows a comparison of this solution with the exact solution. You can see that the collocation method gives a solution that is close to the exact up to $$t = 1$$. However, the same results cannot be used up to $$t = 2$$ without re-evaluating the integrals.

If you think in terms of equation (6) you can see that a weighting function $$v(t)$$ was used to get to the solution. In fact, it is the choice of weighting function that determines whether a method is a collocation method! The weighting function in this case is

{   v(t) = \delta(t-t_j) } $$ where $$t_j$$ is a  node and $$\delta$$ is the Dirac delta function.

Minimizing R(t): Subdomain Method
The subdomain method is another way of minimizing the residuals. In this case, instead of letting the residual vanish at unique points, we let the "average" of the residual vanish over each domain. That is, we let,

\frac{1}{\Delta t_i} \int_{\Delta t_i} R(t) dt = 0 $$ where $$\Delta t_i$$ is the subdomain over which averaging is done. From this definition it is clear that the weighting function for the subdomain method is

v(t) = \begin{cases} 1 & \rm{if} \; t \in \Delta t_i, \\ 0 & \rm{otherwise}. \end{cases} $$

Let us apply the subdomain method to Problem 1. We discretize the domain by choosing one point between $$0$$ and  $$1$$ at $$t = 1/2$$. For the two subdomains (elements) we have,

\frac{1}{\Delta t_1} \int_0^{\frac{1}{2}} R(t) dt = 1 + \frac{5}{4} a_1 + \frac{7}{12} a_2 ~; \qquad\text{and}~\qquad \frac{1}{\Delta t_2} \int_{\frac{1}{2}}^{1} R(t) dt  = 1 + \frac{7}{4} a_1 + \frac{25}{12} a_2~. $$ Setting these residuals to zero and solving for $$a_1$$ and $$a_2$$ we get $$a_1 = -18/19$$ and $$a_2 = 6/19$$. Therefore the approximate solution is

{      u_h(t) = 1 - \frac{18}{19} t + \frac{6}{19} t^2 ~. } $$ Figure 2 shows a comparison of the exact solution and the subdomain and the collocation solutions.

Minimizing R(t): Galerkin Method
In this case, instead of writing our trial function as,

u_h(t) = 1 + a_1 t + a_2 t^2 + \dots + a_n t^n $$ we write it as

u_h(t) = N_0(t) + \sum^n_{i=1} a_i N_i(t) $$ where $$N_0, N_1, \dots, N_n$$ are $$(n+1)$$ linearly independent functions of $$t$$. These are called  basis functions,  interpolation functions, or  shape functions. The first term $$N_0$$ is left outside the sum because it is associated with part or all of the initial or boundary conditions (i.e., we put everything that can be fixed by initial or boundary conditions into $$N_0$$).

Then the trial function in equation (7) can be rewritten using basis functions as

{   u_h(t) = N_0(t) + a_1 N_1(t) + a_2 N_2(t) } $$ where
 * $$\text{(12)} \qquad

{   N_0(t) = 1 ~, N_1(t) = t ~, \text{and} N_2(t) = t^2~. } $$

Important:
In the Galerkin method we choose the basis functions $$N_i~ (i=1 \dots n)$$ as the weighting functions.

If we use $$N_i(t)$$ as the weighting functions $$v(t)$$, equation (6) becomes
 * $$\text{(13)} \qquad

{   \int^1_0 \left(\frac{du_h}{dt} + u_h\right) N_i(t) = 0 \implies \int^1_0 R(t) N_i(t) = 0, \qquad i=1,2~. } $$ Plugging in the value of $$R(t)$$ from equation (8) into equation (13) and using the basis functions from (12) we get

\int^1_0 \left[ 1 + a_1(1+t) + a_2(2t + t^2)\right]~t~dt = 0 \quad\text{and}\quad \int^1_0 \left[ 1 + a_1(1+t) + a_2(2t + t^2)\right]~t^2~dt = 0~. $$ After integrating and solving for $$a_1$$ and $$a_2$$ we get $$a_1 = - 32/35$$ and $$a_2 = 2/7$$. Therefore, the Galerkin approximation we seek is

{ u_h(t) = 1 - \frac{32}{35} t + \frac{2}{7} t^2 }~. $$ Figure 3 shows a comparison of the exact solution with the Galerkin, subdomain, and collocation solutions.

All the approximate solutions diverge from the exact solution beyond $$t = 1$$. The solution to this problem is to break up the domain into  elements so that the trial solution is a good approximation to the exact solution in each element.

Minimizing R(t): Least Squares Method
In the  least-squares method, we try to minimize the residual in a least-squares sense, that is
 * $$\text{(15)} \qquad

\int_0^1 \frac{\partial R(t)^2}{\partial a_i}~dt = \int_0^1 2~R(t)~\frac{\partial R(t)}{\partial a_i}~dt = 0 \equiv {     \int_0^1 R(t)~\frac{\partial R(t)}{\partial a_i}~dt = 0 } $$ where $$a_i = (a_1, a_2)$$. The weighting function for the least squares method is therefore

{   v(t) = \frac{\partial R(t)}{\partial a_i} ~. } $$ Plugging in the value of $$R(t)$$ from equation (8) into equation (15) and using the basis functions from (12) we get

\int^1_0 \left[ 1 + a_1(1+t) + a_2(2t + t^2)\right]~(1+t)~dt = 0 \quad\text{and}\quad \int^1_0 \left[ 1 + a_1(1+t) + a_2(2t + t^2)\right]~(2t+t^2)~dt = 0~. $$ After integrating and solving for $$a_1$$ and $$a_2$$ we get $$a_1 = - 576/611$$ and $$a_2 = 190/611$$. Therefore, the least squares approximation we seek is

{ u_h(t) = 1 - \frac{576}{611} t + \frac{190}{611} t^2 }~. $$ Figure 4 shows a comparison of the exact solution with the Galerkin, subdomain, and collocation solutions.