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Nonlinear finite elements/Nonlinear axial bar weak form

Weak form of non-linear axial bar
Having discussed the equations of the non-linear axial bar, we proceed to construct a finite element approximation. The first step in obtaining the finite element approximation is to construct the weak form. To obtain the weak form we multiply the equilibrium equations discussed in the previous section with a test function (weighting function) and integrate over the whole domain $$[0,L]$$.

\int_{[0,L]}\frac{d}{dx}\left(AE\left(1+\frac{du}{dx}\right)\frac{du}{dx}\right) \varphi +q\varphi=0 $$ $$\varphi$$ is the test function or the weighting function. Now we apply integration by parts to shift a derivative from the first term to the test function.

\int_{[0,L]}AE\left(1+\frac{du}{dx}\right)\frac{du}{dx} \frac{d\varphi}{dx} + \left(AE\left(1+\frac{du}{dx}\right)\frac{du}{dx}\right)|_{x=0}^{x=L}=\int_{[0,L]}q\varphi $$

We consider a pure Dirichlet problem, i.e) no traction conditions are applied on the boundary of the bar, hence we ignore the boundary term. The weak form of the pure Dirichlet problem, for the non-linear axial bar is given by,

\int_{[0,L]}AE\left(1+\frac{du}{dx}\right)\frac{du}{dx} \frac{d\varphi}{dx}=\int_{[0,L]}q\varphi;\;\;\; u(0)=\bar{u} $$ $$\bar{u}$$ is the displacement condition specified on the left boundary. Note that the weak form is not linear in the displacement field $$u$$, this is because of the non-linearity introduced in the constitutive relation. The above equation serves as the starting point for the numerical solution (FEM) of the non-linear axial bar.

Finite element approximation
The weak form of the pure Dirichlet problem for the non-linear axial bar can be thought of as non-linear operators acting on appropriate function space, the problem is to find $$u$$ in an appropriate function space for a prescribed body force $$q$$ such that the weak form is satisfied for all test functions $$\varphi$$. Instead of searching a solution in an infinite dimensional function space, using finite element one approximates the solution in a finite dimensional function space. We denote the finite element approximation for the solution as $$u^h$$ and the finite dimensional approximation for functions in the trial space as $$\varphi^h$$. Here we assume the solution space (trial space) and the test space to be same (spanned by the same set of vectors).

Any function in the test and trial space can be written as,

u^h = \sum_j \tilde{u}_j N_j;\;\;\; \varphi^h =\sum_j \tilde{\varphi}_j N_j $$ Substituting the finite element approximation into the weak form, we get

\int_{[0,L]} AE \left(\left(1+\sum_i \frac{dN_i}{dx}\tilde{u}_i\right)\sum_i \frac{dN_i}{dx}\tilde{u}_i\right)\frac{dN_j}{dx}\tilde{\varphi}_i =     \int_{[0,L]} \sum_jN_j q \tilde{\varphi} $$ Using the arbitrariness of the trial function we can obtain the following relationship between the nodal degrees of freedom and nodal forces.

\int_{[0,L]} AE \left(\left(1+\sum_i \frac{dN_i}{dx}\tilde{u}_i\right)\sum_i \frac{dN_i}{dx}\tilde{u}_i\right)\frac{dN_j}{dx} =     \int_{[0,L]} \sum_jN_j q $$ One should note that the solution for infinite dimensional weak form is now reduced to finding the roots of a non-linear equation in $$\mathbb{R}^n$$. In the next section we will discuss some procedures to solve this non-linear nodal displacement nodal force relation.


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Nonlinear finite elements/Newton Raphson method

The Newton-Raphson method
The Newton-Raphson method is a technique used to find the roots of nonlinear algebraic equations. The method is also called Newton's method.

Let us revisit Newton's method of finding roots in the context of an equation with one degree of freedom. Let the nonlinear equation be

f(x) = 0 ~.\, $$ We wish to find one root of the equation (recall that if $$f(x)\,$$ is linear or quadratic, we can find the root in exact form).

Let us select a point $$x_1\,$$ on the curve (see Figure 1). The slope of the curve at point $$x_1\,$$ is $$df(x_1)/dx\,$$. Therefore, the equation of the tangent to the curve at $$(x_1, f(x_1))\,$$ is

y - f(x_1) = \cfrac{df(x_1)}{dx} (x - x_1) ~. $$ At $$y = 0\,$$, the tangent meets the $$x\,$$-axis at $$x_2\,$$. Therefore we have

-f(x_1) = \cfrac{df(x_1)}{dx} (x_2 - x_1) \qquad \implies \qquad x_2 = x_1 - \cfrac{f(x_1)}{df(x_1)/dx} ~. $$ Recall that the root of $$f(x)\,$$ is the point where the curve of the function meets the $$x\,$$-axis. Therefore, we can repeat the process until we arrive at the root. The Newton iteration scheme is given by

{   x_{r+1} = x_r - \cfrac{f(x_r)}{\cfrac{df(x_r)}{dx}} ~. } $$

The iterative procedure is terminated when the following condition is met:

|x_{r+1} - x_r| < \epsilon \, $$ where $$\epsilon$$ is a small value that depends on the desired accuracy and on machine precision. The Newton-Raphson method gives a quadratic convergence rate. That is, the difference between the answer and the approximate solution is proportional to the previous difference squared. This can be seen by Taylor expanding the non-linear equation and its derivative around the root, and inserting this into the expression for $$ x_{r+1}\,$$.

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


text: File:NewtonIter.png

title: File:NewtonIter.png


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Nonlinear finite elements/Newton method for finite elements

Newton's method for nonlinear finite elements
The finite element approximation of a nonlinear boundary value problem results in a system of nonlinear equations of the form

K(u)u = f $$ Here, K is a nonlinear function relating the nodal degrees of freedom and the internal nodal forces and f is the externally applied nodal forces. Now the problem is to find the values of the nodal degrees of freedom such that the above equation is satisfied. In general the above equation does not have a close form solution (or some algorithm to find the closed form solution), hence we resort to numerical root finding methods to find an approximate solution. Newtons method is commonly used solution procedure for the solution of system simultaneous nonlinear equations. We rewrite the above equation in the form $$g(x) = 0$$

g(u):=K(u) u - f $$ Then, the Newton iteration formula can be written as

\begin{align} u_{r+1} &= u_r - [\cfrac{dg(u_r)}{du}]^{-1} g(u_r) \\ & = u_r - T^{-1}g(u_r) \end{align} $$

The tangent stiffness is given by
 * $$T = \cfrac{dg(u_r)}{du}$$

The iterative procedure is terminated when either the residual is very small or the difference between successive solutions is less than a specified tolerance.

However, both the residual $$\mathbf{r}$$ and the solution $$\mathbf{u}$$ are vectors. We usually compare the $$L_2$$ (Euclidean) norm of the vectors with a tolerance $$\epsilon$$. In symbolic form, we check the norm of the residual using

\lVert\mathbf{r}\rVert_{2} \le \epsilon \qquad \text{or} \qquad \sqrt{\mathbf{r}\cdot\mathbf{r}} \le \epsilon \qquad \text{or} \qquad \sqrt{\sum_{i=1}^n r_i^2} \le \epsilon ~. $$ For the difference between successive solutions we check

\cfrac{\lVert\mathbf{u}_{r+1}-\mathbf{u}_r\rVert_{2}}{\lVert\mathbf{u}_{r+1}\rVert_{2}} \le \epsilon \qquad \text{or} \qquad \cfrac{\sqrt{(\mathbf{u}_{r+1} - \mathbf{u}_r)\cdot(\mathbf{u}_{r+1} - \mathbf{u}_r)}} {\sqrt{\mathbf{u}_{r+1}\cdot\mathbf{u}_{r+1}}} \le \epsilon \qquad \text{or} \qquad \cfrac{\sqrt{\sum_{i=1}^n (u^{r+1}_i - u^r_i)^2}} {\sqrt{\sum_{i=1}^n (u^{r+1}_i)^2}} \le \epsilon ~. $$


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Nonlinear finite elements/Newton method for axial bar

Newton's method applied to the axial bar problem
Let us now return to the axial bar problem and see how the Newton-Raphson scheme can be used to solve the system of equations. The reduced system of equations is

\mathbf{K} \mathbf{u} = \mathbf{f} \qquad \rightarrow \qquad \cfrac{AE_0}{h^2} \begin{bmatrix} (2h-u_3) & -(h+u_2-u_3) \\ -(h+u_2-u_3) & (h+u_2-u_3) \end{bmatrix} \begin{bmatrix} u_2 \\ u_3 \end{bmatrix} = \begin{bmatrix} 0 \\ R \end{bmatrix} ~. $$ Therefore, the residual is

\mathbf{r} = \mathbf{K} \mathbf{u} - \mathbf{f} \qquad \rightarrow \qquad \mathbf{r} = \cfrac{AE_0}{h^2} \begin{bmatrix} (2h-u_3) & -(h+u_2-u_3) \\ -(h+u_2-u_3) & (h+u_2-u_3) \end{bmatrix} \begin{bmatrix} u_2 \\ u_3 \end{bmatrix} - \begin{bmatrix} 0 \\ R \end{bmatrix} ~. $$ The terms of the tangent stiffness matrix are given by

\mathbf{T} := \mathbf{K} + \frac{\partial \mathbf{K}}{\partial \mathbf{u}}~\mathbf{u} \qquad \rightarrow \qquad T_{ij} := K_{ij} + \sum_{m=1}^n \frac{\partial K_{im}}{\partial u_j} u_m ~. $$ For our $$2\times 2$$ reduced stiffness matrix, we have
 * $$\begin{align}

T_{22} & = K_{22} + \frac{\partial K_{22}}{\partial u_2} u_2 + \frac{\partial K_{23}}{\partial u_2} u_3 \\ T_{23} & = K_{23} + \frac{\partial K_{22}}{\partial u_3} u_2 + \frac{\partial K_{23}}{\partial u_3} u_3 \\ T_{32} & = K_{32} + \frac{\partial K_{32}}{\partial u_2} u_2 + \frac{\partial K_{33}}{\partial u_2} u_3 \\ T_{33} & = K_{33} + \frac{\partial K_{32}}{\partial u_3} u_2 + \frac{\partial K_{33}}{\partial u_3} u_3~. \end{align}$$ Now, from the stiffness matrix, we see that
 * $$\begin{align}

K_{22} & = \cfrac{AE_0}{h^2} (2h - u_3) & K_{23} &= -\cfrac{AE_0}{h^2} (h + u_2 - u_3) \\ K_{32} & = -\cfrac{AE_0}{h^2} (h + u_2 - u_3) & K_{33} &= \cfrac{AE_0}{h^2} (h + u_2 - u_3) ~. \end{align}$$ The derivatives are,
 * $$\begin{align}

\frac{\partial K_{22}}{\partial u_2} & = 0 & \frac{\partial K_{23}}{\partial u_2} &= -\cfrac{AE_0}{h^2} & \frac{\partial K_{22}}{\partial u_3} & = -\cfrac{AE_0}{h^2} & \frac{\partial K_{23}}{\partial u_3} &= \cfrac{AE_0}{h^2} \\ \frac{\partial K_{32}}{\partial u_2} &= -\cfrac{AE_0}{h^2} & \frac{\partial K_{33}}{\partial u_2} &= \cfrac{AE_0}{h^2} & \frac{\partial K_{32}}{\partial u_3} &= \cfrac{AE_0}{h^2} & \frac{\partial K_{33}}{\partial u_3} &= -\cfrac{AE_0}{h^2} ~. \end{align}$$ Therefore, the components of $$\mathbf{T}$$ are
 * $$\begin{align}

T_{22} & = \cfrac{AE_0}{h^2} (2h - u_3) + 0 - \cfrac{AE_0}{h^2} u_3 & T_{23} & = -\cfrac{AE_0}{h^2} (h + u_2 - u_3) - \cfrac{AE_0}{h^2} u_2 + \cfrac{AE_0}{h^2} u_3 \\ T_{32} & = -\cfrac{AE_0}{h^2} (h + u_2 - u_3) - \cfrac{AE_0}{h^2} u_2 + \cfrac{AE_0}{h^2} u_3 & T_{33} & = \cfrac{AE_0}{h^2} (h + u_2 - u_3) + \cfrac{AE_0}{h^2} u_2 - \cfrac{AE_0}{h^2} u_3~. \end{align}$$ In matrix form,
 * $$ \text{(8)} \qquad

{   \mathbf{T} = \cfrac{AE_0}{h^2} \begin{bmatrix} 2(h - u_3)        & -h - 2(u_2 - u_3) \\ -h - 2(u_2 - u_3) &  h + 2(u_2 - u_3) \end{bmatrix}~. } $$ The inverse of $$\mathbf{T}$$ (the reduced form shown in equation (8)) is

\mathbf{T}^{-1} = \cfrac{h^2}{AE_0(h - 2u_2)} \begin{bmatrix} 1 & 1 \\          1 & \cfrac{2h - 2u_3}{h + 2u_2 - 2u_3} \end{bmatrix} ~. $$

Maple code
11 A Maple text script for doing these computations symbolically is show below.

Can the tangent stiffness matrix be assembled too?
We have derived the tangent stiffness matrix $$\mathbf{T}$$ from the assembled global stiffness matrix. If the stiffness matrix is large that does not seem to be the way to go. So the question is, can the tangent stiffness matrix be calculated on an element by element basis and assembled just like the global stiffness matrix?

Let us try to figure that out.

The element stiffness matrix has the form

\mathbf{K}^e = \cfrac{AE_0}{h^2}(h + u_1 - u_2)\begin{bmatrix}1 & -1 \\-1 & 1 \end{bmatrix} ~. $$ The tangent stiffness matrix is given by

\mathbf{T}^e = \mathbf{K}^e + \frac{\partial \mathbf{K}^e}{\partial \mathbf{u}}~\mathbf{u}~. $$ After going through the algebra, we get

\mathbf{T}^e = \cfrac{AE_0}{h^2}(h + 2u_1 - 2u_2)\begin{bmatrix}1 & -1 \\-1 & 1 \end{bmatrix} ~. $$ For the two element mesh, the assembled global tangent stiffness matrix is

\mathbf{T} = \cfrac{AE_0}{h^2} \begin{bmatrix} h + 2(u_1 - u_2) & -h - 2(u_1 - u_2) & 0 \\ & 2h + 2(u_1 - u_3) & -h - 2(u_2 - u_3)\\ \text{Symm.}    &                   & h + 2(u_2 - u_3) \\ \end{bmatrix} $$ Applying the boundary condition $$u_1 = 0$$, we get

\mathbf{T} = \cfrac{AE_0}{h^2} \begin{bmatrix} h - 2u_2    & -h + 2u_2  & 0 \\ & 2(h - u_3) & -h - 2(u_2 - u_3)\\ \text{Symm.} &           & h + 2(u_2 - u_3) \end{bmatrix} $$ This matrix is the same as that shown in equation (8). Therefore, we can actually assemble the global tangent stiffness matrix. You could also show that in more general terms. But we avoid that proof here.

Doing the actual computation
The Newton iteration scheme is

\mathbf{u}_{r+1} = \mathbf{u}_r - \mathbf{T}^{-1}(\mathbf{u}_r)~\mathbf{r}(\mathbf{u}_r) $$

Let us go through the process. We will have to start with some assumed values of the parameters.

Let $$A = 0.01$$ m$$^2$$, $$L = 2$$ m, $$E_0 = 100$$ MPa, $$R = 100$$ kN. Since there are two elements, $$h = 1$$ m.

For the first Newton iteration, let the initial guess be

\mathbf{u}_0 = \begin{bmatrix} u_2^0 \\ u_3^0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$ Then, the first Newton iteration gives

\mathbf{K}_0 = 10^6\begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix}, \qquad \mathbf{T}_0 = 10^6\begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix}, \qquad \mathbf{u}_1 = \begin{bmatrix} 0.1 \\ 0.2 \end{bmatrix} ~. $$ The second Newton iteration gives

\mathbf{K}_1 = 10^6\begin{bmatrix} 1.8 & -0.9 \\ -0.9 & 0.9 \end{bmatrix}, \qquad \mathbf{T}_1 = 10^6\begin{bmatrix} 1.6 & -0.8 \\ -0.8 & 0.8 \end{bmatrix}, \qquad \mathbf{u}_2 = \begin{bmatrix} 0.1125 \\ 0.2250 \end{bmatrix} ~. $$ The third Newton iteration gives

\mathbf{K}_2 = 10^6\begin{bmatrix}1.775 & -0.8875\\-0.8875 & 0.8875 \end{bmatrix}, \qquad \mathbf{T}_2 = 10^6\begin{bmatrix}1.55 & -0.775\\-0.775 & 0.775 \end{bmatrix}, \qquad \mathbf{u}_3 = \begin{bmatrix} 0.1127 \\ 0.2254 \end{bmatrix} ~. $$ The fourth Newton iteration gives

\mathbf{K}_3 = 10^6\begin{bmatrix}1.7746&-0.8873\\-0.8873 & 0.8873 \end{bmatrix}, \qquad \mathbf{T}_3 = 10^6\begin{bmatrix}1.5492&-0.7746\\-0.7746 & 0.7746 \end{bmatrix}, \qquad \mathbf{u}_4 = \begin{bmatrix} 0.1127 \\ 0.2254 \end{bmatrix} ~. $$ At this stage we stop the computation because the solution has converged. A plot of the convergence of the Newton iterations in shown in Figure 1. Once we have computed the nodal displacements, we can compute the element stresses in the usual manner. We write the strains as

\varepsilon = \cfrac{du}{dx} = u_1\cfrac{dN_1}{dx} + u_2 \cfrac{dN_2}{dx}~. $$ The stresses are computed using

\sigma = E_0(1-\varepsilon)\varepsilon ~. $$ For the axially loaded bar with no body forces, we have a uniform state of strain (and therefore stress). For the numbers that we have used, these stresses are 10 MPa in each element.

A slightly more complicated situation
At this stage, let us add in the body force terms and work out the solution for another test case.

Let us assume that the input parameters are $$E_0 = 10$$, $$A = 1$$, $$L = 1$$, $$a = 1$$, and $$R = 2$$. The stress-strain curve for this material is shown in Figure 2. We use an error tolerance of $$10^{-6}$$ as a flag to stop the Newton iterations.

Mesh with 10 elements
If we use a mesh containing $$10$$ equally sized elements, we get the solutions shown in Figure 3.

Mesh with 100 elements
If we use a mesh containing $$100$$ equally sized elements, we get the solutions shown in Figure 4.

Even though the displacements and strains are quite different in the linear and the nonlinear cases, the stresses are remarkably exactly the same for the linear and nonlinear cases. This is a result of the form of the modulus that we have chosen to use for this problem. Notice that the error norms show nearly quadratic convergence.

There is however one issue. Due to the shape of the stress-strain curve of the material, Newton iterations do not converge when the applied external load exceeds the amount needed for the strains to exceed 0.5. A method other than Newton-Raphson is needed to solve the problem for such situations.

The Matlab script used for the above calculations is shown below.


text: File:NewtonBar.png

title: File:NewtonBar.png


text: File:SigEpsNLBarNoDim.png

title: File:SigEpsNLBarNoDim.png


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User talk:Pramod kumbhar

Hello Pramod kumbhar, 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! --JWSchmidt 07:03, 15 August 2007 (UTC)

title: User talk:Pramod kumbhar


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Talk:Project management/Introduction

The pictures are very nice, but I don't think they explain much about project management. I'm especially sceptical when there is no mention of documentation or training within the SDLC.TWFred 07:37, 15 August 2007 (UTC)

title: Talk:Project management/Introduction


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Technical writing/Research/SME

"SME" is an abbreviation for "Subject Matter Expert" - a person, who understands the particular problem. One of his or her tasks is to provide the technical writers with pieces of information.

title: Technical writing/Research/SME


text:
BCP/Origanum majorana/Logs

This page contains recent bloom logs for the plant Origanum majorana (Marjoram).

Recent Logs

 * --SB_Johnny | talk 13:11, 7 July 2007 (UTC)
 * --Juan 11:51, 5 October 2007 (UTC)
 * SBJ 19:14, 17 June 2008 (UTC)

=====Archived Logs===== (If there are more than 5 signatures under the recent logs heading, move the oldest logs to under this heading (generally leave the 3 most recent).
 * (no archived logs yet)

title: BCP/Origanum majorana/Logs


text:
Portal talk:Literary Studies/Reading lists/English Literature - Read and discuss

Hello Gilbert, my name is Rachael and I might do, but I am looking for the reading lists and it has brought me here! Just wondering, if you knew where I might find the reading lists Thank you x


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User:Andrew B Clegg~enwikiversity

I am a London-based informatics specialist, currently working on the computation-linguistic analysis of biomedical text.

My personal web page is here.

Andrew B Clegg 12:04, 15 August 2007 (UTC)

title: User:Andrew B Clegg~enwikiversity


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Talk:Jesus

Jesus of Nazareth in Christianity and other religions, is the Firstborn Son of God.
This statement is plainly wrong. Only Christianity states that Jesus is the Firstborn Son of God. Other religions, in particular Islam, accept that Jesus was a prophet sent from God but not that he enjoys any closer relationship with God than any other man due to anything innate. AlistairReece 12:41, 15 August 2007 (UTC)

Use of sources
Please do not quote the Bible to support things which it plainly does not say. The citation from Luke reads as follows (within context):

"In the sixth month, God sent the angel Gabriel to Nazareth, a town in Galilee, to a virgin pledged to be married to a man named Joseph, a descendant of David. The virgin's name was Mary. The angel went to her and said, "Greetings, you who are highly favored! The Lord is with you. Mary was greatly troubled at his words and wondered what kind of greeting this might be. But the angel said to her, "Do not be afraid, Mary, you have found favor with God. You will be with child and give birth to a son, and you are to give him the name Jesus. He will be great and will be called the Son of the Most High. The Lord God will give him the throne of his father David, and he will reign over the house of Jacob forever; his kingdom will never end."

Nowhere is it stated that Jesus was born in the Year 0. Neither does this passage claim deity for Jesus. AlistairReece 12:48, 15 August 2007 (UTC)

 * Re the date of Jesus' birth: When Jesus was born, dates in the Roman Empire were reckoned from the founding of the city of Rome. Later on, after Christianity had become the official religion of the Roman world, the emperor Justinian commissioned a monk named Dionysius Exiguus to make a new calendar which was reckoned from the time of the birth of Christ. This calendar was made in 526 A.D. Unfortunately, long after the Christian calendar was made, it was found that Dionysius had made an error in his calculations and placed the birth of Christ several years later than the actual event. Scholars today generally agree that the most likely date of Christ’s birth is actually 4 or 5 B.C. --gdm 01:02, 11 January 2010 (UTC)

revert
Hello, reverted the last edit. Reason see in comment. Perhaps the author can rewrite in another way ? This seems so far like a doubling:
 * Islam sees Christ as a prophet, but asserts that he is not divine:
 * Islam views Jesus as a great prophet, but nothing more. Erkan Yilmaz Wikiversity:Chat  20:33, 24 February 2008 (UTC)

 * While Jesus is considered a prophet in Islam, it is not really true that he is of the same standing as Muhammed. Muhammed is by far the most important of the prophets because he was the last one, and his prophecies supersede all that came before. Similarly, Muslims recognize the Jewish and Christian scriptures as divinely inspired, but believe they were superseded by the Koran. --gdm 01:25, 11 January 2010 (UTC)

My revisions
I just substantially edited the article, totally reformatting/restructuring it, and adding or rephrasing various details throughout to help clarify. Is this better? The Jade Knight 08:01, 25 March 2008 (UTC)

title: Talk:Jesus


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BCP/Oxalis corniculata/Logs

2007 Logs

 * SBJ 10:27, 25 May 2007 (UTC)
 * SBJ 19:25, 31 May 2007 (UTC)
 * SB Johnny-LM 09:09, 6 June 2007 (UTC)
 * SBJ 16:16, 7 June 2007 (UTC)
 * SBJ 12:47, 15 August 2007 (UTC)
 * SB Johnny-LM 09:33, 8 September 2007 (UTC)
 * SBJ 08:48, 3 October 2007 (UTC)

2008 Logs

 * (no archived logs yet)

Recent Logs

 * --SB_Johnny | talk 10:29, 27 May 2008 (UTC)
 * --SB_Johnny | talk 09:48, 2 July 2008 (UTC)
 * --SB_Johnny | talk 10:21, 14 July 2008 (UTC)
 * Jomegat 23:20, 12 August 2008 (UTC)
 * Jomegat 17:59, 7 October 2008 (UTC)
 * Jomegat 16:31, 10 October 2008 (UTC)

title: BCP/Oxalis corniculata/Logs


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Bloom Clock/Project Discussion

__NEWSECTIONLINK__

This page is intended as a place to discuss the Bloom Clock, as there is much to discuss.

Instruction manuals
I've come up with some instructions for how to use the bcp3 template and how to use DPL to create the keys... currently in my userspace but I'll be moving them now to Bloom Clock/Using Bcp3 and Bloom Clock/Making Keys. I think the rest of the clock is pretty much self-explanatory, though Bcp2 (the one used for unidentified plants) could probably use some improvements as well. If anyone would be willing to improve those instructions a bit, that would be helpful for future contributors. --SB_Johnny | talk 10:48, 3 September 2007 (UTC)

Disambiguation templates
I'd like to come up with some disambiguation templates for genera where the individual species are hard to tell apart. For example, here in PA the goldenrods (Solidago) and many of the Asters (Aster) are almost impossible to tell apart. What the dab template should do is notify users of the issues, and eventually lead to a special key for the genus in question. I might try doing one of these keys for Rudbeckia first, since it's a smaller genus and easier to do. --SB_Johnny | talk 10:59, 3 September 2007 (UTC)
 * I was actually thinking about some similar issues recently. The dab templates sound like a good idea.  What I was wondering was whether we could have multiple pictures of some flowers, showing variants of the same species, on their log pages.  For example, the Torenia in my garden doesn't resemble the picture on the Torenia log page very closely at all, but I know it is the same species.  Particularly with plants they're not familiar with, this kind of thing could confuse some users. --Luai lashire 12:33, 4 September 2007 (UTC)
 * That's part of the reason why the signatures are no longer on the log pages as well in the newer versions of the BCP pages. The idea is to have separate pages for different colors, but send the signatures to the same page. --SB_Johnny | talk 18:14, 4 September 2007 (UTC)

Categories for family and morphology
In order to improve the "Keys" aspect of the clock, I've been thinking we could add some more categories so that more specific DPL pages can be created. The three easiest ones to create would be by family, leaf arrangement, and leaf type. So, for example, you could make a key page for "yellow, daisy-like flowers with simple, alternate leaves that bloom during this time in this region". More categories could be added later as well, to make the searches ever-more accurate. --SB_Johnny | talk 11:33, 3 September 2007 (UTC)
 * If we do that, we should probably have them as sub-pages of the generic keys we already have, or the keys directory pages will clutter up too quickly. --Luai lashire 12:36, 4 September 2007 (UTC)
 * Oh, these would be completely different, in that they'd be smaller, more specific keys (I'll try to make an example later today). For the "disambiguation" keys, I was thinking more of a notification message on the BCP pages that things can get confusing, with a link to the genus key. --SB_Johnny | talk 18:12, 4 September 2007 (UTC)
 * Hmmm... I'm having trouble picturing what you're talking about.  I guess I'll wait for the example, that should make it easier for me to understand.  --Luai lashire 21:49, 5 September 2007 (UTC)

Well, I can't make a morphology one without adding the categories first, but here's an example of a key that's broken down a bit: It would be just as easy to make similar lists for "Herbaceous plants with alternate leaves and white flowers...", etc. --SB_Johnny | talk 10:34, 6 September 2007 (UTC)
 * ../White flowers seen in September in Southeastern Pennsylvania/
 * ../Herbaceous plants with white flowers seen in September in Southeastern Pennsylvania/
 * ../Vines with white flowers seen in September in Southeastern Pennsylvania/
 * ../Shrubs with white flowers seen in September in Southeastern Pennsylvania/
 * I think I see. What I was saying before still aplies, then.  What I meant was that the list of keys would clutter up quickly, so the basic lists should probably still stay "White flowers seen in September in Southeastern PA"; then, when you go to that page, it gives you a breakdown of the catagory- a list of keys like "Vines with white flowers...ect.".  That way navigating is easier, and the list of keys doesn't get too cluttered.  Do you see what I'm saying?  --Luai lashire 22:02, 6 September 2007 (UTC)
 * Yup, definitely with you. The "by color only" keys aren't so huge yet that they need to be broken up anyway, but linking to narrower searches would be good too. I'd like to see it evolve so that you can start with any one reature and then go through the key (i.e., start with "white flower", "herbaceous", "palmate leaves", or "alternate leaves", and it won't matter what order you'll do it in). Ideally we should also be logging fruits by color too, so that a plant could be identified through fruits as well... the bigger it gets the bigger it gets, I suppose. --SB_Johnny | talk 11:41, 9 September 2007 (UTC)
 * BTW: I've started adding family categories (Category:Bloom Clock Plants by Family), thinking about the leaf-related ones now. So far I've thought of these ones:
 * Category:BCP/Alternate leaves
 * Category:BCP/Opposite leaves
 * Category:BCP/Whorled leaves
 * Category:BCP/Basal leaves only
 * Category:BCP/Palmately veined leaves
 * Category:BCP/Pinnately veined leaves
 * Category:BCP/Straplike or bladelike leaves
 * Category:BCP/Simple leaves
 * Category:BCP/Pinnately compound leaves
 * Category:BCP/Palmately compound leaves
 * It's probably a "more's the better" thing, though we can always add more categories later. This might take a few years :). --SB_Johnny | talk 11:46, 9 September 2007 (UTC)
 * Well, then, it's good to get started now, isn't it? :)  They look good so far.
 * As far as logging berries goes, how would that work? Would we have a seperate log page for each plant- one log page for when it's blooming, one for when it's got berries?  --Luai lashire 14:08, 9 September 2007 (UTC)

 * It's so nice to have you around, Luai :). I think we could probably just add a new section on the /Logs pages for fruits, sort of the same way we have been for the bloom logs. New profile pages could then be made for the fruits (assuming we're going to do fruit-specific keys), and have them include both the bloom and fruit logs.
 * OK, that sounds confusing... I'll try just making one for BCP/Cornus amomum (fruiting right now here), to show you what I mean. --SB_Johnny | talk 15:21, 9 September 2007 (UTC)
 * Well, I think I get what you're saying, but that page looks just like a regular bloom page to me. I'm not sure what the difference is.  (sorry I haven't been around, I've been really busy! :)   --Luai lashire 17:58, 15 September 2007 (UTC)

 * Ah, hadn't gotten to it yet, but the fruit data is on there now (as well as a few others). Probably easier to just manually add that in for now, rather then redoing the template or creating a separate log (though a separate log could be handy as well for transcluding to fruit-specific pages). --SB_Johnny | talk 21:54, 17 September 2007 (UTC)
 * Looks good! I think I'll start logging berries myself.  --Luai lashire 20:13, 18 September 2007 (UTC)
 * Yeah, I saw a few more today I need to log (will get to it on the weekend). It would be nice if people using the keys could find out what the berry they saw today was, whether it's edible, etc. --SB_Johnny | talk 23:07, 18 September 2007 (UTC)

switching from categories to templates
To make keystroking and progress tracking easier, I've started switching the leaf morphology categories to templates instead. The templates so far are:

For leaf complexity:
 * 1) bcp/simple leaves
 * 2) bcp/palmately compound leaves
 * 3) bcp/pinnately compound leaves
 * 4) bcp/trifoliate leaves

For leaf arrangement:
 * 1) bcp/alternate leaves
 * 2) bcp/opposite leaves
 * 3) bcp/whorled leaves
 * 4) bcp/basal leaves only

The replacements are being made by a bot. Over the winter I'd like to add more of these for number of petals, leaf shape, etc... using the easier-to-type templates rather than the long-hand category+SUBPAGENAME style that tends to slow down the work. Do you think we should do this for the regional categories as well? --SB_Johnny | talk 12:54, 26 September 2007 (UTC)

Changing subpage form for the keys
To take advantage of the suto-navigation, it would be good to do this, and would also make it easier to move through the keys. For example: moving Bloom Clock/White flowers seen in September in Southeastern Pennsylvania to Bloom Clock/Keys/Southeastern Pennsylvania/September/White Flowers. The page title looks a bit unattractive, but it can be over-ridden using a template. The advantage will work downstream too: for example Bloom Clock/Keys/Southeastern Pennsylvania/October/White Flowers/Herbaceous/Alternate Leaves, etc. --SB_Johnny | talk 11:57, 3 October 2007 (UTC)
 * Aesthetically I guess the new titles don't look great, but they make sense, so I think they're fine. And they do make a lot of sense- much more sense than the old way, especially now that we are adding in the subpages for leaves, ect.  --Luai lashire 20:44, 3 October 2007 (UTC)

Large template change
Hi all. I'll leave messages on everyone's talks later, but with some help from User:Darklama, I've finally made an easy-to-update version of the much-feared bcp3 template. The BCP pages will now just use 3 main embedded templates (+ log pages and smaller templates), rather than the confusing-to-edit substituted version we've been using up until now. The main advantages are that it doesn't use any substituted text, and it is easily updatable by bots running AWB, so we won't have to do things manually any more.

The template is designed to be fairly self-explanatory, but I'll make an instruction manual at Bloom Clock/Creating and editing plant pages. --SB_Johnny | talk 12:45, 26 October 2007 (UTC)

Global warming & urban climates
Is the data of the bloom clock affected by the global warming or urban climates? I not quite sure how to ask the global warming bits, yet I know in urban climates the random and artificial dew schedules may affect bloom times. In the valley I live in, most of the vegetation gets to much hot sun and low humidity to grow. Urban cultures that have planted more trees for shade and installed water sprinklers obviously allows for vegetation to grow even in the hottest summer. In the winter, the artificial dew affects the chill rate on the plants. It's an epic time to find snow in the valley, even when it goes below freezing temp. The dew crystallization kills plants, The lawns gets slight tundra look to them on the top soil, which probably wouldn't happen without artificial dew. Urban climates affects the growth and spread of plants. I don't know if it matters overall if the bloom clock data should show an urban or rural climate for each log entry. Dzonatas 16:02, 6 November 2008 (UTC)

 * Well, the only large data set we have for an urban environment at the moment is from Prague but in general I think people in urban heat islands (or perhaps in the case you describe they would be "shade islands"?) should use a different region tag. Not quite sure though whether your climate would be considered temperate? There was discussion early on about the need for a different kind of keying for regions where the dominant factor is wet/dry rather than warm/cold.
 * I think we'd need a much larger data set to catch global warming data (both in terms of years logged and of regions logged from), and we'd likely need a finer-grained data set as well (I imagine it would take a considerable amount of warming before we saw changes that would show up by month). The problem with trying to get a finer grain is that it would require people to log a lot more often, which might make a chore out of a hobby :-). --SB_Johnny talk 18:21, 6 November 2008 (UTC)

 * I understand now. I missed any emphasis on the hobby aspect. It was kinda funny how much I thought about this. I'm not an expert here, but I have some basic horticulture knowledge. I thought yesterday about somehow to classify plants as if they are on a sun, moon, star, or wild schedule as a way to avoid climate details. Now I read it more and I see you are doing something likewise. Dzonatas 18:48, 7 November 2008 (UTC)

 * Oh definitely, the fun part of the clock is having fun engaging with it... I'm a professional horticulturist but never enjoyed taking note of the first or last flower as much as I have since working on this. Have a look at Jomegat's page for an example of it's "life changing potential". I've also been much more motivated learn the names and life cycles of weeds and wildflowers, which has proven to be rather fruitful professionally. --SB_Johnny talk 23:42, 7 November 2008 (UTC)

title: Bloom Clock/Project Discussion


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Portal talk:International Business

Hello! My name is Sunnie and I work for MSUglobal at Michigan State University. We currently have resources on international business that we are putting up in WikiEducator. Would this content be something that we could put here in the International Business section? Our goal in putting up this free content is to collaborate with people on a global scale and learn how they localize and repurpose this information. It would help us in learning about ways business is done in other countries.

Please let me know if this is something that we should pursue. Thanks! MSUglobal, 13:49, 15 August 2007

title: Portal talk:International Business


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User talk:Miss Bellatrix

Hello Miss Bellatrix, 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:
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Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --JWSchmidt 14:12, 15 August 2007 (UTC)

title: User talk:Miss Bellatrix


text:
User talk:MSUglobal

Hello MSUglobal, 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:
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international business
"We currently have resources on international business" <-- I assume that means you have learning resources. Learning resources of all types are welcome. It sounds like your main goal might be "learning about ways business is done in other countries". Wikiversity is also open to research projects. --JWSchmidt 14:47, 15 August 2007 (UTC)

Image copyright problems

 * The currently affected images are listed below. For each, please provide a license template and source. Thanks. – Mike.lifeguard | @en.wb 23:15, 13 February 2008 (UTC)

 * 2007-09-25 14:58, MSUglobal: Wikiversitylogo.png: untagged! ; 0 (0)
 * 2007-09-25 16:53, MSUglobal: Globewikiversity.jpg: untagged! ; 263 (239)
 * 2007-09-25 18:46, MSUglobal: World Unit 1.2.jpg: untagged! ; 1 (1)
 * 2007-09-25 19:16, MSUglobal: Chart Unit 3.4.jpg: untagged! ; 1 (1)
 * 2007-09-28 15:49, MSUglobal: Bank Reference Form.pdf: untagged! ; 1 (1)
 * 2007-09-28 15:53, MSUglobal: Credit Application.pdf: untagged! ; 1 (1)
 * 2007-09-28 15:54, MSUglobal: Vendor Reference Form.pdf: untagged! ; 1 (1)
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 * 2007-09-28 16:01, MSUglobal: International Documents Listing.pdf: untagged! ; 1 (1)
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Google Analytics
Hi. I see you added a Google Analytics tracker to Trade Finance recently. Another user didn't understand what it was and removed it. My guess is that the hit tracker wouldn't work like that anyway (did it work?!). Inserting or trying to insert external scripts into Wikiversity pages probably isn't kosher, but if you'd like to experiment with such things, I suggest you bring up the topic on the colloquium first where people who know more than I do may be able to advise you. --McCormack 06:11, 26 August 2008 (UTC)

title: User talk:MSUglobal


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User:Kbdank71

I'm Kbdank71. I'm an admin at the English Wikipedia.

My other user pages:
 * Wikipedia: w:User:Kbdank71
 * Meta: m:User:Kbdank71
 * Wiktionary: wikt:User:Kbdank71
 * Commons: commons:User:Kbdank71
 * Wikisource: s:User:Kbdank71
 * Wikiquote: q:User:Kbdank71
 * Wikinews: n:User:Kbdank71
 * Wikibooks: b:User:Kbdank71
 * Wikispecies wikispecies:User:Kbdank71
 * Wikiversity User:Kbdank71

title: User:Kbdank71


text:
User talk:Trappe

Hello Trappe, 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:
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Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --JWSchmidt 15:30, 15 August 2007 (UTC)

Stanford Dragon
Hello, Trappe! We cannot use renders of the Stanford Dragon on Wikipedia or the Commons because the model, and depictions of it, are under noncommercial licenses. However, your photograph I believe avoids that by being a depiction of a derivative work, a separate entity. Could you upload your photograph to the Commons so that a reference image will be publicly available? LokiClock 00:24, 3 June 2010 (UTC)

title: User talk:Trappe


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User talk:Jazzy-devine

Hello Jazzy-devine, 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:
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Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --JWSchmidt 15:31, 15 August 2007 (UTC)

title: User talk:Jazzy-devine


text:
Markerless Tracking/Performance

Performance is a very important factor when it comes to tracking objects. To gain interactive or even real-time update rates it's necessary to soften the above stated assumptions in the most harmless way (see accuracy below). In the following you could read and work on possibly good ways to speed up the whole process.

Mathematical optimization of the function space
Using optimization algorithms like adaptive random search or Particle Filters are as far as known by the authors a very safe way to prevent the system form simulating all possible configurations without loosing accuracy.

Naturally the optimization will be harder if the degrees of freedom increases. But as it has been shown by Deutscher et. al. that it is possible to optimize high dimensional functions with so called annealed particle filtering. This seems to be very easy archivable when using images. Just take a lower resolution and you will get a smoothed out function. Nontheless it need to be discussed about possible errors and accuracy of such an aproach.

"Little changes" assumption
If we assume only little changes between two captured frames we could dramatically reduce the search space of possible real world configurations.

Using specialized hardware to generate hypothesis
Simulating possible captured datasets on specialized hardware is definitely a very good thing to do. Especially for images this easily done by using modern graphic cards.

title: Markerless Tracking/Performance


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Markerless Tracking/Accuracy

It's possible to derive some conclusions about the accuracy from the assumptions stated on the main research project page.

capturing
The accuracy of the system is only as good as the data captured from reality. For example it's not possible to detect fully occluded objects with a normal camera. Also a wrong focal distance or bad lighting will hinder good tracking.

generating
The accuracy of the system is only as good as the data generated by the computer. For example it's just a rough assumption of real light conditions to use a single one point light and standard OpenGL based rendering.

Note: The simulation must include the whole evolutionary story of the image. If an camera captures images with a radial distorition, the renderer should do the same.

selecting
The accuracy of the system is only as good as the mechanism which compares generated and captured datasets. Completly identical data must result in 100% similarity. That's easy. The tricky part is to get a good statement about almost mathching and partial matching data. See the selection page for a discussion.

title: Markerless Tracking/Accuracy


text:
User talk:Vitor

Hello Vitor, 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;
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 * 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 16:46, 15 August 2007 (UTC)

Thanks
Super, thanks for the examples! --HappyCamper 17:10, 17 August 2007 (UTC)

title: User talk:Vitor


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Partial differential equations

Partial differential equations (PDEs) are the most common method by which we model physical problems in engineering. Finite element methods are one of many ways of solving PDEs. This handout reviews the basics of PDEs and discusses some of the classes of PDEs in brief. The contents are based on  Partial Differential Equations in Mechanics volumes 1 and 2 by A.P.S. Selvadurai and Nonlinear Finite Elements of Continua and Structures by T. Belytschko, W.K. Liu, and B. Moran.

Definition of a PDE
A PDE is a relationship between an unknown function of several variables and its partial derivatives.

Let $$u(x_1, x_2, x_3, t)$$ be an unknown function. The  independent variables are $$x_1$$, $$x_2$$, $$x_3$$, and $$t$$. We usually write

u = u(x_1, x_2, x_3, t) $$ and say that $$u$$ is the  dependent variable.

Partial derivatives are denoted by expressions such as

u_{,1} = \frac{\partial u}{\partial x_1} ~; u_{,2} = \frac{\partial u}{\partial x_2} ~; u_{,11} = \frac{\partial^2 u}{\partial x_1\partial x_1} \equiv \frac{\partial^2 u}{\partial x_1^2} ~; u_{,12} = \frac{\partial^2 u}{\partial x_1\partial x_2}~. $$

Some examples of partial differential equations are
 * $$\begin{align}

u_{,t} = u_{,1} + u_{,2} &\Leftrightarrow \frac{\partial u}{\partial t} = \frac{\partial u}{\partial x_1} + \frac{\partial u}{\partial x_2} \\ \nabla^2 u = 0 \Leftrightarrow u_{,11} + u_{,22} + u_{,33} = 0 &\Leftrightarrow \frac{\partial^2 u}{\partial x_1^2} + \frac{\partial^2 u}{\partial x_2^2} + \frac{\partial^2 u}{\partial x_3^2} = 0 \\ u_{,1111} = u_{,22} + u &\Leftrightarrow \frac{\partial^4 u}{\partial x_1^4} = \frac{\partial^2 u}{\partial x_2^2} + u    ~. \end{align}$$

An example of a system of partial differential equations is

\boldsymbol{\nabla} (\boldsymbol{\nabla} \bullet \mathbf{u}) + \nabla^2 \mathbf{u} + \mathbf{f} = \mathbf{0} \Leftrightarrow u_{k,ki} + u_{i,jj} + f_i = 0 $$ In expanded form this system of equations is
 * $$\begin{align}

\frac{\partial^2 u_1}{\partial x_1^2} + \frac{\partial^2 u_2}{\partial x_2\partial x_1} + \frac{\partial^2 u_3}{\partial x_3\partial x_1} + \frac{\partial^2 u_1}{\partial x_1^2} + \frac{\partial^2 u_1}{\partial x_2^2} + \frac{\partial^2 u_1}{\partial x_3^2} + f_1 & = 0 \\   \frac{\partial^2 u_1}{\partial x_1\partial x_2} + \frac{\partial^2 u_2}{\partial x_2^2} + \frac{\partial^2 u_3}{\partial x_3\partial x_2} + \frac{\partial^2 u_2}{\partial x_1^2} + \frac{\partial^2 u_2}{\partial x_2^2} + \frac{\partial^2 u_2}{\partial x_3^2} + f_2 & = 0 \\   \frac{\partial^2 u_1}{\partial x_1\partial x_3} + \frac{\partial^2 u_2}{\partial x_2\partial x_3} + \frac{\partial^2 u_3}{\partial x_3^2} + \frac{\partial^2 u_3}{\partial x_1^2} + \frac{\partial^2 u_3}{\partial x_2^2} + \frac{\partial^2 u_3}{\partial x_3^2} + f_3 & = 0  \end{align}$$

It is often more convenient to write PDEs in vector notation or index notation.

Order of a PDE
The order of a PDE is determined by the highest derivative in the equation. For example,
 * $$\begin{align}

\frac{\partial u}{\partial t} - \frac{\partial u}{\partial x} & = 0 ~\text{is a first-order PDE.}\\ \frac{\partial^2 u}{\partial x_1^2} + \frac{\partial^2 u}{\partial x_2^2} + \frac{\partial^2 u}{\partial x_3^2} & = 0 ~\text{is a second-order PDE.}\\ \frac{\partial^4 u}{\partial x_1^4} + \frac{\partial^2 u}{\partial x_2^2} - u & = 0 ~\text{is a fourth-order PDE.}\\ \left(\frac{\partial u}{\partial x_1}\right)^3 + \frac{\partial u}{\partial x_2} + u^4 & = 0 ~\text{is a first-order PDE.} \end{align}$$

Linear and nonlinear PDEs
A  linear PDE is one that is of first degree in all of its field variables and partial derivatives. For example,
 * $$\begin{align}

\frac{\partial u}{\partial x_1} + \frac{\partial u}{\partial x_2} & = 0 \text{is linear}~.\\ \frac{\partial u}{\partial x_1} + \left(\frac{\partial u}{\partial x_2}\right)^2 & = 0 \text{is nonlinear}~.\\ \frac{\partial u}{\partial x_1} + \frac{\partial u}{\partial x_2} + u^2 & = 0 \text{is nonlinear}~.\\ \frac{\partial^2 u}{\partial x_1^2} + \frac{\partial^2 u}{\partial x_2^2} & = x_1 \text{is linear}~.\\ \frac{\partial^2 u}{\partial x_1^2} + u\frac{\partial^2 u}{\partial x_2^2} & = 0 \text{is quasilinear}~. \end{align}$$

The above equations can also be written in operator notation as
 * $$\begin{align}

D(u) = 0 & \text{where} D := \frac{\partial }{\partial x_1} + \frac{\partial }{\partial x_2}~. \\   D(u) = 0 & \text{where} D := \frac{\partial }{\partial x_1} + \left(\frac{\partial }{\partial x_2}\right)^2~.\\ D(u) = 0 & \text{where} D := \frac{\partial }{\partial x_1} + \frac{\partial }{\partial x_2} + u^2~.\\ D(u) = x_1 & \text{where} D := \frac{\partial^2 }{\partial x_1^2} + \frac{\partial^2 }{\partial x_2^2}~.\\ D(u) = 0 & \text{where} D := \frac{\partial^2 }{\partial x_1^2} + u\frac{\partial^2 }{\partial x_2^2}~. \end{align}$$

Homogeneous PDEs
Let $$L$$ be a linear operator. Then a linear partial differential equation can be written in the form

L(u) = f(x_1,x_2,x_3,t)~. $$ If $$f(x_1,x_2,x_3,t) = 0$$, the PDE is called homogeneous. For example,
 * $$\begin{align}

\frac{\partial u}{\partial t} + \frac{\partial u}{\partial x_1} + \frac{\partial u}{\partial x_2} + \frac{\partial u}{\partial x_3} & = 0 \text{is homogeneous}~.\\ \frac{\partial u}{\partial t} + \frac{\partial u}{\partial x_1} + \frac{\partial u}{\partial x_2} + \frac{\partial u}{\partial x_3} & = x_1 + x_2 \text{is nonhomogeneous}~.\\ \end{align}$$

Elliptic, Hyperbolic, and Parabolic PDEs
We usually come across three-types of second-order PDEs in mechanics. These are classified as  elliptic,  hyperbolic, and  parabolic.

The equations of elasticity (without inertial terms) are  elliptic PDEs.  Hyperbolic PDEs describe wave propagation phenomena. The heat conduction equation is an example of a  parabolic PDE.

Each type of PDE has certain characteristics that help determine if a particular finite element approach is appropriate to the problem being described by the PDE. Interestingly, just knowing the type of PDE can give us insight into how smooth the solution is, how fast information propagates, and the effect of initial and boundary conditions.
 * In hyperbolic PDEs, the smoothness of the solution depends on the    smoothness of the initial and boundary conditions.  For instance, if     there is a jump in the data at the start or at the boundaries, then the jump will propagate as a discontinuity in the solution.  If, in addition, the    PDE is  nonlinear, then shocks may develop even though the initial    conditions and the boundary conditions are smooth.  In a system modeled with a hyperbolic PDE, information travels at a finite speed referred to as the wavespeed.  Information is not transmitted until the wave arrives.
 * In contrast, the solutions of elliptic PDEs are always smooth, even if the initial and boundary conditions are rough (though there may be singularities at sharp corners). In addition, boundary data     at any point affect the solution at all points in the domain.
 * Parabolic PDEs are usually time dependent and represent the diffusion-like processes. Solutions are smooth in space but may possess singularities.  However, information travels at infinite speed in a parabolic system.

Suppose we have a second-order PDE of the form

a(x_1,x_2) \frac{\partial^2 u}{\partial x_1^2} + b(x_1,x_2) \frac{\partial^2 u}{\partial x_1\partial x_2} + c(x_1,x_2) \frac{\partial^2 u}{\partial x_2^2} + d(x_1,x_2) \frac{\partial u}{\partial x_1} + e(x_1,x_2) \frac{\partial u}{\partial x_2} + f(x_1,x_2) u = g(x_1,x_2) $$

Then, the PDE is called  elliptic if

{    b^2 - 4ac < 0 } $$ An example is

\frac{\partial^2 u}{\partial x_1^2} + \frac{\partial^2 u}{\partial x_1\partial x_2} + \frac{\partial^2 u}{\partial x_2^2} = x_1 \frac{\partial u}{\partial x_1} $$

The PDE is called  hyperbolic if

{    b^2 - 4ac > 0 } $$ An example is

\frac{\partial^2 u}{\partial x_1^2} + 3\frac{\partial^2 u}{\partial x_1\partial x_2} + \frac{\partial^2 u}{\partial x_2^2} = x_1 \frac{\partial u}{\partial x_1} $$

The PDE is called  parabolic if

{    b^2 - 4ac = 0 } $$ An example is

\frac{\partial^2 u}{\partial x_1^2} + 2\frac{\partial^2 u}{\partial x_1\partial x_2} + \frac{\partial^2 u}{\partial x_2^2} = x_1 \frac{\partial u}{\partial x_1} $$

Solutions to Common PDEs
Partial differential equation appear in several areas of physics and engineering. A firm grasp of how to solve ordinary differential equations is required to solve PDEs. In particular, solutions to the Sturm-Liouville problems should be familiar to anyone attempting to solve PDEs.

 * A tutorial on how to solve the Laplace equation
 * A tutorial on how to solve the Poisson equation
 * A tutorial on how to apply the method of separation of variables

Application of PDEs in Physics and Engineering
There are many applications of partial differential equations in physics and engineering. Here are some examples:
 * The diffusion equation
 * The heat equation
 * The wave equation
 * Maxwell's equations

Resources

 * Numerical partial differential equations
 * MIT 18.03 Differential Equations, Spring 2006

Équation différentielle

The Heat conduction equation of 2-D is elliptic in space and parabolic in time.

title: Partial differential equations


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Markerless Tracking/Realization

On of the main targets of our research project Markerless Tracking is to implement the stated idea for the use in Robotics and Augmented Reality. Because the aim is quite clear but the best way of realization is unsettled we are using a prototype-centered software development approach.

Prototype A1

The first prototype was developed by Dipl.-Inform Rodja Trappe, Dipl.-Inform. Matthias Dennhardt and Dipl.-Inform Tobias Feldmann at the University of Koblenz.

Setting
While capturing a physical version of the Stanford dragon (archived by rapid prototyping) with a single 640x480 fire-wire camera in a controlled light situation an OpenGL Rendering-Engine generates a whole bunch of images with possible locations of the dragon.

Workflow
After capturing an image the adaptive random search-optimization is started to obtain the relative pose between camera and dragon. The optimization algorithm is therefor able to evaluate a 6D function. This evaluation is implemented as an similarity measure between captured image and the on-the-fly generated computer graphics image. The measure is based on the well known normalized cross-correlation.

Technical Details
Prototype A is written in C++ and only compiles on Linux Systems. Ogre was used as Rendering-Engine and unicap for image aquisition. To load, manipulate and handle images we used the DevIL image library.

Implementation Details
The rendering was done with 512x512 resolution and completely downloaded into cpu memory for computing the similarity between generated and captured image.

Results
We were able to find the pose of the dragon program in about 70 seconds. We switched into an other optimization mode to keep track of little movements after first pose was found. Now it was possible to turn and move the dragon (~1cm) each ten seconds without loosing track of the object.

Prototype A2
The second prototype is heavily based on the first one and targets the same setting with th e same workflow. It was written by Dipl.-Inform Rodja Trappe to repair some bugs that had been occured, to gain more speed and archive greater robustness in pose estimation.

comparing images
The similarity mesure beween rendering and capturing was improoved by not normalizing the colors with the mean of the image. Also the performance could be increased.

The C++ implementation is as follows:

compareImages(RGBAImage* imageA, RGBAImage* imageB){ float distance; int pixelCount = 0; CGImage::byte* dataA = imageA->getRaw; CGImage::byte* dataB = imageB->getRaw; for(int i = 0; i < size; i++){ // save the color channels Vec3 cA(*dataA++,*dataA++,*dataA++); Vec3 cB(*dataB++,*dataB++,*dataB++); // if we have no alpha: compute distance of colors and sum up the result if (*dataA != 0){ pixelCount++; distance += (cA - cB).length; } 		dataA++; dataB++; } 	Vec3 meanDistance = distance / pixelCount; float difference = meanDistance / MAX_DISTANCE_IN_COLORSPACE; float similarity = 1 - difference; return similarity; }

With MAX_DISTANCE_IN_COLORSPACE beeing the constant $$\sqrt{3} * 255$$.

Prototype A3
The first prototype was enhanced by implementing the image comperation on the GPU. This results in a current speedup to 10 ms per hypothesis. The CPU implementation took about 12 ms for image download from the graphicscard and 10 ms for the image comperation. With the new implementation it is possible to have nearly interactive framerates after initial detection of the pose.

Prototype B
Note: This prototype is currently in planing phase.

An adaptive random search-optimization as it has been implemented in Prototype A seems to be very slow converging. The used implementation also badly adapts when the function values get bad over time. Therefor we would like to implement a Particle Filter to replace the old optimization algorithm.

title: Markerless Tracking/Realization


text: File:Tux Paint Resources Folder.png

title: File:Tux Paint Resources Folder.png


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Nonlinear finite elements/Lagrangian and Eulerian descriptions

In our study of the axially loaded bar, we have assumed that the area of cross section remains constant during the deformation. Therefore, the physical stress is not affected by the area of cross section of the rod. However, in most materials, the area of cross section changes with deformation. This change is most prominent during  large deformations. In this handout, we will discuss a  Lagrangian finite element formulation for large deformations.

There are two main ways of approaching problems that involve the motion of deformable materials - the  Lagrangian way and the  Eulerian way. These approaches are distinguished by three important aspects:

 * 1) The mesh description.
 * 2) The stress tensor and momentum equation (kinetics).
 * 3) The strain measure (kinematics).

Definitions
Some essential definitions are given below:

 *  Spatial or Eulerian coordinates ($$\mathbf{x}$$):
 * These coordinates are used to locate a point in space with respect to a fixed basis. You can think of these coordinates as the ones you are familiar with.

 *  Material or Lagrangian coordinates ($$\boldsymbol{X}$$):
 * These coordinates are used to  label material points. If we sit on a material point, the label does not change with time. We do start with a  reference label which we usually choose as the initial spatial coordinates of a material point.

 *  Motion or Deformation  $$(\boldsymbol{\varphi}(\boldsymbol{X},t))$$:
 * A motion or deformation is defined as a mapping between the initial and the current configuration. We usually write this relationship as

\mathbf{x} = \boldsymbol{\varphi}(\boldsymbol{X}, t) ~. $$
 * The initial configuraton is

\boldsymbol{X} = \boldsymbol{\varphi}(\boldsymbol{X}, 0) ~. $$
 *  Displacement  $$(\mathbf{u}(\boldsymbol{X},t))$$:
 * The displacement is defined as the difference between the reference and the current configuration. We write

\mathbf{u}(\boldsymbol{X}, t) = \boldsymbol{\varphi}(\boldsymbol{X}, t) - \boldsymbol{X} \equiv \mathbf{x} - \boldsymbol{X} ~. $$
 *  Velocity  $$(\mathbf{v}(\boldsymbol{X},t))$$:
 * The velocity of a material point is the derivative of the motion with $$\boldsymbol{X}$$ fixed.

\mathbf{v}(\boldsymbol{X}, t) = \frac{\partial }{\partial t}(\boldsymbol{\varphi}(\boldsymbol{X}, t)) ~. $$
 *  Acceleration  $$(\mathbf{a}(\boldsymbol{X},t))$$:
 * The acceleration of a material point is the derivative of the velocity with $$\boldsymbol{X}$$ fixed.

\mathbf{a}(\boldsymbol{X}, t) = \frac{\partial }{\partial t}(\mathbf{v}(\boldsymbol{X}, t)) ~. $$

An example
Let us consider a one-dimensional problem. Assume that the motion is

x = \varphi(X, t) = 1 + X(1 + 2t + t^2) ~. $$ The inverse of the map $$\varphi$$ gives us $$X$$ in terms of $$x$$, i.e.,

X = \varphi^{-1}(x, t) = \cfrac{x - 1}{1 + 2t + t^2} ~. $$ Then the displacement of the material point $$X$$ is

u(X, t) = x - X = 1 + X(2t + t^2) ~. $$ The velocity of the material point is

v(X, t) = \frac{\partial x}{\partial t} = 2X(1 + t) \qquad \leftarrow \qquad \text{ Lagrangian description of velocity.} $$ We could alternatively wish to express the velocity in terms of the spatial coordinate $$x$$. In that case, we have

\bar{v}(x, t) = v(\varphi^{-1}(x,t),t) = \cfrac{2(x-1)(1+t)}{1+ 2t + t^2} \qquad \leftarrow \qquad \text{ Eulerian description of velocity.} $$ Note that $$v(X,t)$$ and $$\bar{v}(x,t)$$ give us different ways of expressing the same quantity. To keep the number of symbols reasonably small, we usually represent both functions with the same symbol ($$v$$) and determine which form we are talking about on the basis of the arguments.

We can find the  Lagrangian and  Eulerian versions of the acceleration in a similar manner. Thus,

a(X, t) = \frac{\partial v}{\partial t} = 2X \qquad\leftarrow \qquad \text{Lagrangian description of acceleration,} $$ and

a(x, t) = a(\varphi^{-1}(x,t),t) = \cfrac{2(x-1)}{1+ 2t + t^2} \qquad\leftarrow \qquad \text{Eulerian description of acceleration.} $$

Lagrangian and Eulerian Meshes
The Lagrangian and Eulerian descriptions can be visualized in terms of the corresponding meshes (see Figure 1).

We can think of the Lagrangian mesh as being drawn on the body. The mesh deforms with the body. Both the nodes and the material points change position as the body deforms. However, the position of the material points relative to the nodes remains fixed.

On the other hand, the Eulerian mesh is a background mesh. The body flows through the mesh as it deforms. The nodes remain fixed and the materials points move through the mesh. The position of a material point relative to the nodes varies with the motion.

Some features, advantages, and disadvantages of the two descriptions are given below.

 Lagrangian mesh
 * 1) Lagrangian coordinates of nodes move with the material. Material coordinates of material points are time invariant.
 * 2) No material passes between elements.
 * 3) Element quadrature points remain coincident with material points.
 * 4) Boundary nodes remain on the boundary. Therefore, boundary conditions and interface conditions are easily applied.
 * 5) Severe mesh distortion can occur because the mesh deforms with the material.

 Eulerian mesh
 * 1) Eulerian coordinates of nodes are fixed and coincide with spatial points. Spatial coordinates of material points vary with time.
 * 2) Material flows through the mesh.
 * 3) The material point at a given element quadrature point changes with time. This makes dealing with history-dependent materials difficult.
 * 4) Boundary nodes and the material boundary may not coincide. Therefore, boundary conditions and interface conditions are hard to apply.
 * 5) There is no mesh distortion because the mesh is fixed in space. However, the domain that needs to be modeled is larger because we do not want the body to leave the domain.


text: File:LagrangianMesh.png

title: File:LagrangianMesh.png


text: File:EulerianMesh.png

title: File:EulerianMesh.png


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File:Noblerinthemind Seduced by the dark side eg.pdf

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


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Nonlinear finite elements/Lagrangian finite elements

Lagrangian finite elements
Two types of approaches are usually taken when formulating Lagrangian finite elements:

 * 1)  Total Lagrangian:
 * 2) *The stress and strain measures are Lagrangian, i.e.,they are defined with respect to the original configuration.
 * 3) *Derivatives and integrals are computed with respect to the Lagrangian (or material) coordinates $$(\mathbf{X})$$.
 * 4)  Updated Lagrangian:
 * 5) *The stress and strain measures are Eulerian, i.e.,they are defined with respect to the current configuration.
 * 6) *Derivatives and integrals are computed with respect to the Eulerian (or spatial) coordinates $$(\mathbf{x})$$.

The following 1-D examples illustrate what these approaches entail.

Consider the axially loaded bar shown in Figure 1. In the reference (or initial) configuration, the bar has a length $$L_0$$, an area $$A_0(X)$$, and density $$\rho_0(X)$$. A tensile force $$T$$ is applied at the free end. In the current (or deformed) configuration at time $$t$$, the length of the bar increases to $$L$$, the area decreases to $$A(X, t)$$, and the density changes to $$\rho(X,t)$$.

Motion in Lagrangian form
The  motion is given by

{ x = \varphi(X, t) = x(X, t) ~, \qquad X \in [0, L_0]~. } $$

For the reference configuration,

X = \varphi(X, 0) = x(X, 0) ~. $$

The  displacement is

{ u(X, t) = \varphi(X, t) - X = x - X ~. } $$

For the reference configuration,

u_0 = u(X, 0) = \varphi(X, 0) - X = X - X = 0 ~. $$

The  deformation gradient is

{ F(X, t) = \frac{\partial }{\partial X}[\varphi(X,t)] = \frac{\partial x}{\partial X} ~. } $$

For the reference configuration,

F_0 = F(X, 0) = \frac{\partial }{\partial X}[\varphi(X,0)] = \frac{\partial X}{\partial X} = 1 ~. $$

The  Jacobian determinant of the motion is (regarding this step, read the Discuss page)

{ J = \cfrac{A}{A_0} F ~. } $$

For the reference configuration,

J_0 = \cfrac{A_0}{A_0} F_0 = 1 ~. $$


text: File:TotalLagrangian.png

title: File:TotalLagrangian.png


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Nonlinear finite elements/Motion in Lagrangian form

Motion in Lagrangian form
The  motion is given by

{ x = \varphi(X, t) = x(X, t) ~, \qquad X \in [0, L_0]~. } $$ For the reference configuration,

X = \varphi(X, 0) = x(X, 0) ~. $$

The  displacement is

{ u(X, t) = \varphi(X, t) - X = x - X ~. } $$ For the reference configuration,

u_0 = u(X, 0) = \varphi(X, 0) - X = X - X = 0 ~. $$

The  deformation gradient is

{ F(X, t) = \frac{\partial }{\partial X}[\varphi(X,t)] = \frac{\partial x}{\partial X} ~. } $$ For the reference configuration,

F_0 = F(X, 0) = \frac{\partial }{\partial X}[\varphi(X,0)] = \frac{\partial X}{\partial X} = 1 ~. $$ The  Jacobian determinant of the motion is

{ J = \cfrac{A}{A_0} F ~. } $$ For the reference configuration,

J_0 = \cfrac{A_0}{A_0} F_0 = 1 ~. $$


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Nonlinear finite elements/Total Lagrangian approach

Total Lagrangian Approach
In the total Lagrangian approach, the discrete equations are formulated with respect to the reference configuration. The independent variables are $$X$$ and $$t$$. The dependent variable is the displacement $$u(X,t)$$.

Total Lagrangian Stress and Strain Measures
The  strain is

{ \varepsilon(X, t) = F(X,t) - 1 = \frac{\partial x}{\partial X} - 1 = \frac{\partial u}{\partial X} = u_{,X} ~. } $$ For the reference configuration,

\varepsilon_0 = \varepsilon(X, 0) = F(X,0) - 1 = \frac{\partial X}{\partial X} - 1 = 0 ~. $$ The  Cauchy stress (force/current area) is

{ \sigma = \cfrac{T}{A} ~. } $$ The  engineering stress (force/initial area) is

{ P = \cfrac{T}{A_0} ~. } $$ The two stresses are related by

\sigma = \cfrac{A_0}{A} P ~; \qquad P = \cfrac{A}{A_0} \sigma ~. $$ For the reference configuration,

\sigma = P ~. $$

Governing Equations in Total Lagrangian Form
The  Total Lagrangian governing equations are:

 *  Conservation of Mass:

{ \rho~J = \rho_0~J_0 = \rho_0~. } $$
 * For the axially loaded bar,

\rho\cfrac{A}{A_0} F = \rho_0 \qquad \implies \qquad \rho~A~F = \rho_0~A_0 ~. $$
 *  Conservation of Momentum:

{ \frac{\partial }{\partial X}(A_0~P) + \rho_0~A_0~b = \rho_0~A_0~\frac{\partial^2 u}{\partial t^2} ~. } $$
 * For the bar $$A_0$$ is constant. Therefore,

\frac{\partial P}{\partial X} + \rho_0~b = \rho_0~\frac{\partial^2 u}{\partial t^2} ~. $$
 * In short form,

P_{,X} + \rho_0~b = \rho_0~\ddot{u} ~. $$
 * If $$\ddot{u} \approx 0$$, we get the  equilibrium equation

P_{,X} + \rho_0~b = 0 ~. $$
 *  Conservation of Energy:
 * For the bar:

{ \rho_0\frac{\partial W_{\text{int}}}{\partial t} = \frac{\partial F}{\partial t}~P ~. } $$
 * In short form:

\rho_0\dot{W}_{\text{int}} = \dot{F}~P ~. $$
 *  Constitutive Equations:
 *  Total Form:
 * For a linear elastic material:

{ P(X,t) = E^{PF}~\varepsilon(X,t) = E^{PF}~u_{,X} = E^{PF}\left[F(X,t) - 1\right]~. } $$
 * The superscript $$PF$$ refers to the fact that this function relates $$P$$ and $$F$$.
 * For small strains:

E^{PF} = \text{Youngs modulus}~. $$
 *  Rate Form:
 * For a linear elastic material:

{ \dot{P}(X,t) = E^{PF}~\dot{\varepsilon}(X,t) = E^{PF}\dot{F}(X,t)~. } $$

The Wave Equation
The momentum equation is

P_{,X} + \rho_0~b = \rho_0~\ddot{u} ~. $$ The total form constitutive equation is

P(X,t) = E^{PF}~\varepsilon(X,t) = E^{PF}~u_{,X} ~. $$ Plug constitutive equation into momentum equation to get

\left(E^{PF}~u_{,X}\right)_{,X} + \rho_0~b = \rho_0~\ddot{u} ~. $$ If $$E^{PF}$$ is constant in the bar and the body force is zero,

{ E^{PF}~u_{,XX} = \rho_0~\ddot{u} \qquad \implies \qquad u_{,XX} = \cfrac{\rho_0}{E^{PF}}~\ddot{u} = \cfrac{1}{c_0^2}~\ddot{u}~. } $$ This is the wave equation (hyperbolic PDE). The wave speed is $$c_0$$. If acceleration is zero, the equation becomes elliptic.

Initial and Boundary Conditions
The governing equation for the rod is second-order in time. So two initial conditions are needed.

{ \begin{align} u(X,0) & = u_0(X) & ~\text{for}~ & X \in [0, L] \\ \dot{u}(X,0) & = v_0(X) & ~\text{for}~ & X \in [0, L] \end{align} } $$ The rod is initially at rest. Therefore, the initial conditions are

\begin{align} u(X,0) & = 0 & ~\text{for}~ & X \in [0, L] \\ \dot{u}(X,0) & = 0 & ~\text{for}~ & X \in [0, L] \end{align} $$

Since the problem is one-dimensional, there are two boundary points. Also, the momentum equation is second-order in the displacements. Therefore, at each end, either $$u$$ or $$u_{,X}$$ must be prescribed. In mechanics, instead of $$u_{,X}$$, the traction is prescribed.

Let $$\Gamma_u$$ be the part of the boundary where displacements are prescribed. Let $$\Gamma_t$$ be the part of the boundary where tractions (force vector per unit area) are prescribed.

Then the  displacement boundary conditions are

{ u(X,t) = \bar{u}(X,t) \qquad ~\text{for}~X \in \Gamma_u ~. } $$ The  traction boundary conditions are

{ n^0~P(X,t) = t^0_X(X,t) \qquad ~\text{for}~X \in \Gamma_t ~. } $$ The  unit normal to the boundary in the reference configuration is $$n^0$$.

For the axially loaded bar, the displacement boundary condition is

u(0,t) = 0 \qquad ~\text{for}~X = 0 $$ The unit normal to the boundary is

\begin{align} n^0 & = -1 & ~\text{at}~ & X = 0 \\ n^0 & = 1& ~\text{at}~ & X = L \end{align} $$ Therefore, the traction boundary condition is

P(L,t) = T(t) \qquad ~\text{for}~X = L ~. $$ For shock problems or fracture problems, an addition interior continuity or  jump condition may be needed. This condition is written as

{ \lfloor P\rceil = 0 } $$ where

\lfloor P(X)\rceil = P(X+\epsilon) - P(X-\epsilon)~, \qquad \epsilon \rightarrow 0 ~. $$

Weak Form for Total Lagrangian
The momentum equation (in its full form) is

(A_0 P)_{,X} + \rho_0 A_0 b = \rho_0 A_0 \ddot{u} ~. $$ To get the weak form over an element, we multiply the equation by a weighting function and integrate over the length of the element (from $$X_a$$ to $$X_b$$).

\int_{X_a}^{X_b} w\left[(A_0 P)_{,X} + \rho_0 A_0 b - \rho_0 A_0 \ddot{u}\right]~dX = 0 ~. $$ Integrate the first term by parts to get

\int_{X_a}^{X_b} w (A_0 P)_{,X}~dX = \left[w A_0 P\right]_{X_a}^{X_b} - \int_{X_a}^{X_b} w_{,X} (A_0 P)~dX ~. $$ Plug the above into the weak form and rearrange to get

\int_{X_a}^{X_b} w_{,X} (A_0 P)~dX + \int_{X_a}^{X_b} w\rho_0 A_0 \ddot{u}~dX = \int_{X_a}^{X_b} w \rho_0 A_0 b~dX +\left[w A_0 P\right]_{X_a}^{X_b} ~. $$ If we think of the weighting function $$w$$ as a variation of $$u$$ that satisfies the kinematic admissibility conditions, we get

{ \int_{X_a}^{X_b} (\delta u)_{,X} (A_0 P)~dX + \int_{X_a}^{X_b} \delta u\rho_0 A_0 \ddot{u}~dX = \int_{X_a}^{X_b} \delta u \rho_0 A_0 b~dX +\left[\delta u A_0 P\right]_{X_a}^{X_b} ~. } $$ Recall that

u = \varphi(X,t) - X ~. $$ Therefore,

\delta u = \delta \varphi(X,t) = \delta x ~. $$ Taking a derivative of this variation, we have

\delta u_{,X} = \delta x_{,X} = \delta F ~. $$ Also,

\begin{align} \left[\delta u A_0 P\right]_{X_a}^{X_b} &= [\delta u A_0 P]_{X_b} - [\delta u A_0 P]_{X_a} \\ &=[\delta u A_0 P n_0]_{X_b} + [\delta u A_0 P n_0]_{X_a} \\ &=[\delta u A_0 t^0_X]_{X_b} + [\delta u A_0 t^0_X]_{X_a} \\ &= \left[\delta u A_0 t^0_X\right]_{\Gamma_t} ~. \end{align} $$ Therefore, we can alternatively write the weak form as

{ \int_{X_a}^{X_b} \delta F A_0 P~dX + \int_{X_a}^{X_b} \delta u\rho_0 A_0 \ddot{u}~dX = \int_{X_a}^{X_b} \delta u \rho_0 A_0 b~dX +\left[\delta u A_0 t^0_X\right]_{\Gamma_t}~. } $$ Comparing with the energy equation, we see that the first term above is the internal virtual work and the weak form is the principle of virtual work for the 1-D problem. The weak form may also be written as

{ \delta W = \delta W_{\text{int}} - \delta W_{\text{ext}} + \delta W_{\text{kin}} = 0 } $$ where,

{ \begin{align} \delta W_{\text{int}} & = \int_{X_a}^{X_b} (\delta u)_{,X} (A_0 P)~dX \\ \delta W_{\text{ext}} & = \int_{X_a}^{X_b} \delta u \rho_0 A_0 b~dX + \left[\delta u A_0 P\right]_{X_a}^{X_b} \\ \delta W_{\text{kin}} & = \int_{X_a}^{X_b} \delta u\rho_0 A_0 \ddot{u}~dX \end{align} } $$

Finite Element Discretization for Total Lagrangian
The  trial solution is

{ u_h(X,t) = \sum_{j=1}^n N_j(X) u_j(t) } $$ where $$n$$ is the number of nodes. In matrix form,

{ u_h(X,t) = \mathbf{N}~\mathbf{u} } \text{where} \mathbf{N} = \begin{bmatrix} N_1(X) & N_2(X) && N_n(X) \end{bmatrix} ~\text{and}~ \mathbf{u} = \begin{bmatrix} u_1(t) \\ u_2(t) \\ \vdots \\ u_n(t) \end{bmatrix} ~. $$ The  test (weighting) function is

{ \delta u_h(X) = \sum_{i=1}^n N_i(X) \delta u_i ~. } \text{or} { \delta u_h(X) = \mathbf{N}~\delta \mathbf{u} } \text{where} \delta \mathbf{u} = \begin{bmatrix} \delta u_1 \\ \delta u_2 \\ \vdots \\ \delta u_n \end{bmatrix} ~. $$ The  derivatives of the test functions with respect to $$X$$ are

{ (\delta u_h)_{,X} = \sum_{i=1}^n N_{i,X} \delta u_i ~. } \text{or} { (\delta u_h)_{,X} = \mathbf{B}_0~\delta \mathbf{u} } \text{where} \mathbf{B}_0 = \begin{bmatrix} N_{1,X} & N_{2,X} && N_{n,X} \end{bmatrix} ~. $$ We will derive the finite element system of equations after substituting these into the approximate weak form

\int_{X_a}^{X_b} (\delta u_h)_{,X} (A_0 P)~dX + \int_{X_a}^{X_b} \delta u_h\rho_0 A_0 \ddot{u}_h~dX = \int_{X_a}^{X_b} \delta u_h \rho_0 A_0 b +\left[\delta u_h A_0 P\right]_{X_a}^{X_b} ~. $$ Let us proceed term by term.

First LHS Term
The first term represents the  virtual internal work

\delta W_{\text{int}} = \int_{X_a}^{X_b} (\delta u_h)_{,X} (A_0 P)~dX ~. $$ Plugging in the derivative of the test function, we get

\begin{align} \delta W_{\text{int}} & = \int_{X_a}^{X_b} \left[\sum_{i=1}^n N_{i,X}\delta u_i\right] (A_0 P)~dX\\ & = \sum_{i=1}^n \delta u_i \left[\int_{X_a}^{X_b} N_{i,X} (A_0 P)~dX\right]\\ & = \sum_{i=1}^n \delta u_i f_i^{\text{int}} ~. \end{align} $$ The last substitution is made because the virtual internal work is the work done by internal forces in moving through a virtual displacement. The matrix form of the expression for virtual internal work is

{ \delta W_{\text{int}} = \delta \mathbf{u}^T \mathbf{f}_{\text{int}} } \text{where} \mathbf{f}_{\text{int}} = \begin{bmatrix} f_1^{\text{int}} \\ f_2^{\text{int}} \\ \vdots \\ f_n^{\text{int}} \end{bmatrix} ~. $$ The internal force is

{ f_i^{\text{int}} = \int_{X_a}^{X_b} N_{i,X} (A_0 P)~dX } \text{or} { \mathbf{f}_{\text{int}} = \int_{X_a}^{X_b} \mathbf{B}_0^T~(A_0 P)~dX = \int_{\Omega_0} \mathbf{B}_0^T~P~d\Omega_0~. } $$

Second LHS Term
The second term represents the  virtual kinetic work

\delta W_{\text{kin}} = \int_{X_a}^{X_b} \delta u_h\rho_0 A_0 \ddot{u}_h~dX ~. $$ Plugging in the test function, we get

\begin{align} \delta W_{\text{kin}} & = \int_{X_a}^{X_b} \left[\sum_{i=1}^n \delta u_i N_i\right] \rho_0 A_0 \ddot{u}_h~dX\\ & = \sum_{i=1}^n \delta u_i \left[\int_{X_a}^{X_b} \rho_0 A_0 N_i \ddot{u}_h~dX\right]\\ & = \sum_{i=1}^n \delta u_i f_i^{\text{kin}} ~. \end{align} $$ The inertial (kinetic) force is

{ f^{\text{kin}}_i =\int_{X_a}^{X_b} \rho_0 A_0 N_i \ddot{u}_h~dX ~. } $$ Now, plugging in the trial function into the expression for the inertial force, we get

\begin{align} f^{\text{kin}}_i & = \int_{X_a}^{X_b} \rho_0 A_0 N_i \left[\sum_{j=1}^n \ddot{u}_j N_j\right]~dX \\ & = \sum_{j=1}^n \left[\int_{X_a}^{X_b} \rho_0 A_0 N_i N_j~dX\right] \ddot{u}_j \\ & = \sum_{j=1}^n M_{ij} a_j \end{align} $$ where

{ M_{ij} := \int_{X_a}^{X_b} \rho_0 A_0 N_i N_j~dX \qquad \leftarrow \quad \text{Consistent Mass Matrix Term.} } $$ and

a_j := \ddot{u}_j \qquad\leftarrow \quad\text{Acceleration.} $$ Note that the mass matrix is  independent of time!

In matrix form,

{ \mathbf{f}_{\text{kin}} = \mathbf{M} \mathbf{a} } \text{where} \mathbf{a} = \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{bmatrix} = \begin{bmatrix} \ddot{u}_1 \\ \ddot{u}_2 \\ \vdots \\ \ddot{u}_n \end{bmatrix} \text{and} \mathbf{f}_{\text{kin}} = \begin{bmatrix} f_1^{\text{kin}} \\ f_2^{\text{kin}} \\ \vdots \\ f_n^{\text{kin}} \end{bmatrix} ~. $$ The consistent mass matrix in matrix notation is

{ \mathbf{M} =\int_{X_a}^{X_b} \rho_0 A_0 \mathbf{N}^T \mathbf{N} ~dX = \int_{\Omega_0} \rho_0~\mathbf{N}^T~\mathbf{N}~d\Omega_0 ~. } $$ Plugging the expression for the inertial force into the expression for virtual kinetic work, we get

\begin{align} \delta W_{\text{kin}} & = \sum_{i=1}^n \delta u_i \left[\sum_{j=1}^n M_{ij} a_j\right] \\ & = \sum_{i=1}^n \sum_{j=1}^n \delta u_i M_{ij} a_j \end{align} $$ In matrix form,

{ \delta W_{\text{kin}} = (\delta \mathbf{u})^T \mathbf{M} \mathbf{a} = (\delta \mathbf{u})^T~\mathbf{f}_{\text{kin}} ~. } $$

RHS Terms
The right hand side terms represent the  virtual external work

\delta W_{\text{ext}} = \int_{X_a}^{X_b}\delta u_h\rho_0 A_0 b~dX + \left[\delta u_h A_0 t^0_X\right]_{\Gamma_t}~. $$ Plugging in the test function into the above expression gives

\begin{align} \delta W_{\text{ext}} & = \int_{X_a}^{X_b}\left[\sum_{i=1}^n N_i \delta u_i\right]\rho_0 A_0 b~dX + \left[\left(\sum_{i=1}^n N_i \delta u_i\right) A_0 t^0_X\right]_{\Gamma_t} \\ & = \sum_{i=1}^n \delta u_i \left[\int_{X_a}^{X_b} N_i \rho_0 A_0 b~dX\right] + \left[\sum_{i=1}^n \delta u_i N_i A_0 t^0_X\right]_{\Gamma_t} \\ & = \sum_{i=1}^n \delta u_i \left[\int_{X_a}^{X_b} N_i \rho_0 A_0 b~dX + \left[N_i A_0 t^0_X\right]_{\Gamma_t}\right] \\ & = \sum_{i=1}^n \delta u_i~f^{\text{ext}}_i ~. \end{align} $$ In matrix notation,

{ \delta W_{\text{ext}} = \delta \mathbf{u}^T~ \mathbf{f}_{\text{ext}} } \text{where} \mathbf{f}_{\text{ext}} = \begin{bmatrix} f_1^{\text{ext}} \\ f_2^{\text{ext}} \\ \vdots \\ f_n^{\text{ext}} \end{bmatrix} ~. $$ The external force is given by

{ f_i^{\text{ext}} =\int_{X_a}^{X_b} N_i \rho_0 A_0 b~dX + \left[N_i A_0 t^0_X\right]_{\Gamma_t}~. } $$ In matrix notation,

{ \mathbf{f}_{\text{ext}} = \int_{X_a}^{X_b} \mathbf{N}^T \rho_0 A_0 b~dX + \left[\mathbf{N}^T A_0 t^0_X\right]_{\Gamma_t} }\text{or} { \mathbf{f}_{\text{ext}} = \int_{\Omega_0} \rho_0~\mathbf{N}^T~b~d\Omega_0 + \left[\mathbf{N}^T A_0 t^0_X\right]_{\Gamma_t}~. } $$

Collecting the terms, we get the finite element system of equations

\sum_{i=1}^n \delta u_i \left[f_i^{\text{int}} + f_i^{\text{kin}} - f_i^{\text{ext}}\right] = 0 $$ Now, the weighting function is arbitrary except at nodes where displacement BCs are prescribed. At these nodes the weighting function is zero.

For the bar, let us assume that a displacement is prescribed at node $$1$$. Then, the finite element system of equations becomes

f_i^{\text{int}} + f_i^{\text{kin}} - f_i^{\text{ext}} = 0 \qquad \text{for} i = 2 \dots n $$ or,

{ \sum_{j=1}^n M_{ij}~a_j + f_i^{\text{int}} - f_i^{\text{ext}} = 0 \qquad \text{for} i = 2 \dots n ~. } $$ Since the displacement at node $$1$$ is known, the acceleration at node $$1$$ is also known. Note that we have to differentiate the displacement twice to get the acceleration and hence the prescribed displacement must be a $$C^1$$ function.

We can take the known acceleration $$a_1$$ to the RHS to get

\sum_{j=2}^n M_{ij}~a_j + f_i^{\text{int}} - f_i^{\text{ext}} = -M_{i1}~a_1 \qquad \text{for} i = 2 \dots n ~. $$ The above equation shows that prescribed boundary accelerations (and hence prescribed boundary displacements) contribute to nodes which are not on the boundary.

We can avoid that issue by making the mass matrix diagonal (using ad-hoc procedures that conserve momentum). If the mass matrix is diagonal, we get

M_{ii}~a_i + f_i^{\text{int}} - f_i^{\text{ext}} = 0 \qquad \text{for} i = 2 \dots n ~. $$ In matrix form,

{ \mathbf{M}~\mathbf{a} = \mathbf{f}_{\text{ext}} - \mathbf{f}_{\text{int}} } \text{or} { \mathbf{f} = \mathbf{M}~\mathbf{a}. }~. $$ One way of generating a  diagonal mass matrix or  lumped mass matrix is the row-sum technique. The rows of the mass matrix are summed at lumped at the diagonal of the matrix. Thus,

\begin{align} M^{\text{diagonal}}_{ii} & = \sum_{j=1}^n M^{\text{consistent}}_{ij} \\ & = \sum_{j=1}^n \int_{X_a}^{X_b} \rho_0 N_i N_j A_0 ~dX \\ & = \int_{X_a}^{X_b} \rho_0 N_i \left[\sum_{j=1}^n N_j\right] A_0 ~dX \\ & = \int_{X_a}^{X_b} \rho_0 N_i A_0 ~dX\qquad\text{since}~\sum_{j=1}^n N_j = 1~. \end{align} $$

Discrete Equations for Total Lagrangian
The finite element equations in total Lagrangian form are:

{ \begin{align} u(X,t) & = \sum_i N_i(X)~u_i^e(t) = \mathbf{N}~\mathbf{u}^e \\ \varepsilon(X,t) & = \sum_i N_{i,X}~u_i^e(t) = \mathbf{B}_0~\mathbf{u}^e \\ F(X,t) & = \sum_i N_{i,X}~x_i^e(t) = \mathbf{B}_0~\mathbf{x}^e \\ \mathbf{f}^e_{\text{int}} & = \int_{\Omega_0^e} \mathbf{B}_0^T P~d\Omega_0 \\ \mathbf{f}^e_{\text{ext}} & = \int_{\Omega_0^e} \rho_0 \mathbf{N}^Tb~d\Omega_0 + \left[\mathbf{N}^T A_0 t_X^0 \right]_{\Gamma_t^e} \\ \mathbf{M}^e & = \int_{\Omega_0^e} \rho_0 \mathbf{N}^T \mathbf{N} ~d\Omega_0 \\ \mathbf{M}~\ddot{\mathbf{u}} & = \mathbf{f}_{\text{ext}} - \mathbf{f}_{\text{int}} ~. \end{align} } $$