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Psycholinguistics/Hemispheric Lateralization of Language

Introduction
Hemispheric lateralization refers to the distinction between functions of the right and left hemispheres of the brain. If one hemisphere is more heavily involved in a specific function, it is often referred to as being dominant (Bear et al., 2007). Lateralization is of interest with regards to language, as it is believed that language is a heavily lateralized function: certain aspects of language are found to be localized in the left hemisphere, while others are found in the right, with the left hemisphere most often dominant. This was initially proposed by early lesion-deficit models and studies with split-brain patients, and has been shown in more recent years through tests like the Wada test and imaging studies. There have been studies which show that there are anatomic asymmetries located near and around the regions associated with language, and each hemisphere has shown to play its own but separate role in the production and comprehension of speech. The hemispheric lateralization of language functions has been suggested to be associated with both handedness, sex, bilingualism, sign-language, and a variance amongst cultures. It has also been proposed that a reorganization occurs following brain injury that involves a shifting of lateralized function, as long as the injury occurs early in life.

Jean Baptiste Bouillaud and Simon Alexandre Ernest Aubertin
French physician Jean Baptiste Bouillaud (1796-1881) was one of the earliest proponents of hemispheric language lateralization. On February 21, 1825, Bouillaud presented a paper to the Royal Academy of Medicine in France which suggested that, because so many human tasks are performed using the right hand (such as writing), the left hemisphere might be the in control of that hand. This observation implies that language, at the core of writing, would be localized in the left hemisphere. It was already known at this time that motor function was primarily controlled by the hemisphere ipsilateral to the side of the body through lesion studies. Bouillaud also proposed that speech is localized in the frontal lobes, a theory that was carried on by Bouillaud’s son-in-law Simon Alexandre Ernest Aubertin (1825-1893), who went on to work with famed French neurologist Paul Broca in 1861. Together, Aubertin and Broca examined a patient with a left frontal lobe lesion who had lost nearly all ability to speak; this case and several others similar to it became the basis behind the earliest theories of language lateralization.

Paul Broca
French neurologist Paul Broca (1824-1880) is often credited as being the first to expound upon this theory of language lateralization. In 1861, a 51-year-old patient named Leborgne came to Broca; Leborgne was almost completely unable to speak and suffered from cellulitis of the right leg. Leborgne was able to comprehend language but was mostly unable to produce it. He responded to almost everything with the word “tan” and thus came to be known as Tan. Broca theorized that Tan must have a lesion of the left frontal lobe, and this theory was confirmed in autopsy when Tan died later that year (Bear et al., 2007). In 1863, Broca published a paper in which he described eight cases of patients with damage to the left frontal lobe, all of whom had lost their ability to produce language, and included evidence of right frontal lesions having little effect on articulate speech (Bear et al., 2007). These findings led Broca to propose, in 1864, that the expression of language is controlled by a specific hemisphere, most often the left (Bear et al., 2007). “On parle avec hemisphere gauche,” Broca concluded (Purves et al., 2008)- we speak with the left hemisphere.

Carl Wernicke
German anatomist Carl Wernicke (1848-1904) is also known as an early supporter of the theory of language lateralization. In 1874, Wernicke found an area in the temporal lobe of the left hemisphere, distinct from that which Broca had described, which disrupted language capabilities (Bear et al., 2007). He then went on to provide the earliest map of left hemisphere language organization and processing.

Lesion Studies
A good deal of what we know about language lateralization comes from studying the loss of language abilities following brain injury (Bear et al., 2007). Aphasia, the partial or complete loss of language abilities occurring after brain damage, is the source of much of the information on this subject (Bear et al., 2007). As shown in the studies of Bouillaud, Aubertin, Broca and Wernicke described above, lesion studies combined with autopsy reports can tell us a a lot about the localization of language, which ultimately has supplied information on lateralization. Lesion studies have shown that, not only is the left cerebral hemisphere most often dominant for language, but also that the right hemisphere generally is not, as lesions in the right hemisphere rarely disturb speech and language function (Bear et al., 2007).

The dangers of using lesion studies are, of course, that they may overemphasize the relevance of particular localized areas and their associated functions. The connection between brain regions and behaviours is not always simple, and is often based on a larger network of connections. This is shown in the fact that the severity of an individual’s aphasia is often related to the amount of tissue damaged around the lesion itself (Bear et al., 2007). It is also known that there is a difference in the severity of the deficit depending on whether the area was removed surgically, or was caused by stroke. This is the case because strokes affect both the cortex and the subcortical structures; this is due to the location of the middle cerebral artery, which supplies blood to the areas associated with language, as well as involvement of the basal ganglia, and is often the cause of stroke. As such, surgically produced lesions tend to have milder effects than those resulting from stroke (Bear et al., 2007).

]

Split Brain Studies
Studies of patients who have had commissurotomies (split-brain patients) have provided significant information about language lateralization. Commissurotomy is a surgical procedure in which the hemispheres are disconnected by cutting the corpus callosum, the massive bundle of 200 million axons connecting the right and left hemisphere (Bear et al., 2007). Following this procedure, almost all communication between the hemispheres is lost, and each hemisphere then acts independently of the other. What is striking about split-brain patients with regards to the study of language lateralization is that a word may be presented to the right hemisphere of a patient whose left hemisphere is dominant, and when the patient is asked to name the word they will say that nothing is there. This is because, although the right hemisphere “saw” the word, it is the left hemisphere which “speaks.” If that same word is presented to the left hemisphere, the patient is able to verbalize the response (Bear et al., 2007). As such, split-brain patients have presented substantial evidence that language function is generally lateralized in the left hemisphere.

Wada test
The Wada test was created by Juhn Wada at the Montreal Neurological Institute in 1949, and was designed specifically to study lateralization. A fast-acting barbiturate such as sodium amytal is injected into the carotid artery on one side (although current procedures prefer to use a catheter which is inserted into the femoral artery), and is then transported to the cerebral hemisphere on the opposite side. It then serves to anaesthetize that side of the brain for approximately 10 minutes, after which it begins to wear off and the functions which were disrupted by the anaesthetic gradually return, often displaying aphasic errors (Bear et al., 2007; Wada and Rasmussen, 1960). During the time in which the patient is anaesthetized, he or she is assessed on their ability to use language. If the left hemisphere is anaesthetized and is the dominant hemisphere, the patient loses all ability to speak, whereas if the left hemisphere is anaesthetized but the right hemisphere is dominant, the patient will continue to speak throughout the procedure (Bear et al., 2007).

In a study published in 1977, Brenda Milner used the Wada test to demonstrate that 98% of right-handed people and 70% of left-handed people have a dominant left hemisphere with regards to language and speech function. Her results also showed that 2% of right-handed people have a dominant right hemisphere, which is the same percentage of patients that display aphasia following a lesion to the right hemisphere (Branch et al., 1964).

This procedure is also used prior to brain surgery in order to determine the dominant hemisphere, so as to avoid removal of an area associated with speech and language.

Functional transcranial Doppler ultrasonography
Functional transcranial Doppler ultrasonography (fTCD) is a non-invasive method for examining event-related changes in cerebral blood flow velocity in the middle cerebral arteries(Knecht et al., 1998). This technique can reliably assess which hemisphere is dominant and to what extent, which regards to language lateralization. Studies using fTCD have shown a linear relationship between handedness and language (Knecht et al., 2000).

Electrical stimulation, TMS and Imaging
Electrical stimulation was pioneered by Wilder Penfield and his colleagues at the Montreal Neurological Institute in the 1930s, and helped to identify certain lateralized areas associated with speech and language. Electrical stimulation is the application of an electrical current directly to the cortical tissue of a patient who is conscious. Penfield found that stimulating the left frontal or temporal regions of the left hemisphere with an electrical current accelerated the production of speech. He also found that stimulation can cause inhibition in complex functions like language, as applying a current to the areas associated with speech production in the left hemisphere while the patient is engaged in speech serves to disrupt this behaviour (Penfield, 1963). This procedure is performed during surgery while the skull is removed, and as such it is not a commonly used method of assessment.

Transcranial Magnetic Stimulation (TMS) is a non-invasive procedure, often combined in studies with MRI, which has helped to map the regions associated with speech, showing lateralization to be dominant in the left hemisphere. TMS has also shown that, following brain injury, it is more likely that it is the tissue surrounding the lesion that acts in a compensatory way rather than the opposite hemisphere providing compensation. The major drawback of TMS is, of course, the fact that the magnetic stimulation must pass through the scalp, skull, and meninges before stimulating the brain region of choice.

Imaging studies have proven to be incredibly useful in determining lateralization of language abilities. Functional magnetic resonance imaging (fMRI) and positron emission tomography (PET) have been able to show the complex circuitry associated with speech and language; they have also proven to be consistent with the findings from previous lesion studies, as well as Penfield’s electric stimulation (Bear et al., 2007). There has been some controversy regarding bilateral activation shown in fMRI studies, the reasons unknown, however it has been suggested that perhaps the right hemisphere is involved in aspects of speech that are not measured by such tests as the Wada procedure (Bear et al., 2007). A significant finding is that fMRI results during developmental years show activation during speech and the use of language mainly in the left hemisphere, providing further evidence in support of left hemisphere dominance (Bear et al., 2007).

Cerebral Dominance: Language Functions of The Left and Right Hemispheres
The perisylvian cortex of the left hemisphere is involved in language production and comprehension, which is why it is often referred to as dominant, or said to "speak" (Ojemann, G. A., 1991; Purves et al., 2008). Roger Sperry and his colleagues’ split-brain studies have shown that the left hemisphere is also responsible for lexical and syntactic language (grammatical rules, sentence structure), writing and speech (Purves at al., 2008). Other aspects of language which are thought to be governed in most people by the left hemisphere include audition of language-related sounds, recognition of letters and words, phonetics and semantics.

The right hemisphere, though generally not dominant in terms of linguistic ability, has its role in the use of language. Split-brain studies present evidence that, despite the right hemisphere having no “speech,” it is still able to understand language through the auditory system. It also has a small amount of reading ability and word recognition. Lesion studies of patients who have right hemisphere lesions show a reduction in verbal fluency and deficits in the understanding and use of prosody. Patients who have had their right hemisphere surgically removed (hemispherectomy) show no aphasia, but do show less obvious deficiencies in areas such as verbal selection and understanding of metaphor. It has thus been concluded that the right hemisphere is most often responsible for the prosodic and emotional elements of speech and language (Purves et al., 2008).

Anatomical Asymmetries
The structural differences between the right and left hemisphere may play a role in the lateralization of language. In the nineteenth century, anatomists observed that the left hemisphere’s Sylvian fissure (lateral sulcus) is longer and less steep than that of the right (Bear et al., 2007). In 1980, Graham Ratcliffe and his colleagues used evidence of this asymmetry of the Sylvian fissure, shown in carotid angiogram, combined with results of Wada testing, and found that individuals with speech regions located in the left hemisphere had a mean difference of 27 degrees in the angle of the blood vessels leaving the posterior end of the Sylvian fissure, while those with language located in the right hemisphere had a mean angle of zero degrees.

] In the 1960s, Norman Geschwind and his colleagues at Harvard Medical School found that the planum temporale, the superior portion of the temporal lobe, is larger in the left hemisphere in almost two thirds of humans (Geschwind & Levitsky, 1968), an observation which was later confirmed with MRI (Bear et al., 2007; Purves et al., 2008). This asymmetry exists even in the brain of the human fetus (Bear et al., 2007). The correlation of this asymmetry with the left hemisphere’s language dominance is refuted by many due to the fact that 67% of people show this structural asymmetry, while 97% show left hemispheric dominance. Another problem which exists in examining asymmetry of the planum temporale is how the anterior and posterior borders of this region are defined, and the fact that investigators differ in this definition. This is especially a problem when the transverse gyrus of Heschl, used to mark the anterior of the planum temporale, appear in double (which is not unusual). There are differing opinions as to whether or not the second transverse gyrus should be defined as being within the planum temporale, or outside of it (Beardon, A. A., 1997).

Handedness
The correlation between handedness and hemispheric lateralization is described in the results of the Wada test, described above. The majority of the population is right handed (approximately 90%), and the Wada test results propose that 93% of people’s left hemisphere is dominant for language (Bear et al., 2007). A linear relationship between handedness and langage has been shown using fTCD in a study done by Knecht et al. (2008); their findings show an 27% incidence for right hemisphere dominance in their group of left-handers, a finding consistent with the notion of there being a linear relationship between handedness and incidence of right hemisphere dominance in left-handers (Knecht et al., 2000). This study used a word generation task, and admits that perhaps a measurement of prosody or other such suspected right hemisphere functions may have a different relationship with handedness (Knecht et al., 2000). It is also true that correlation does not necessarily imply causation, and it is also suggested that there is no direct relationship between handedness and language at all, as the majority of left-handers also have their language lateralized in the left hemisphere (Purves et al., 2008). It is, however, a physical example of functional asymmetry, and it is certainly possible that a more substantial connection between handedness and language will be found.

Gender Differences
The tendency for women to score higher than men on language-related tasks is perhaps the result of the fact that women also tend to have a larger corpus callosum than men, indicating more neural connections between the right and left hemispheres. fMRI studies show that women have more bilateral activation than men when performing rhyming tasks, and PET studies show that women have more bilateral activation than men during reading tasks. Perhaps the bilateral activation implies the use of what are thought to be right hemisphere language abilities, such as prosody and intonation. Research has also shown that women have a greater ability to recover from left hemisphere brain damage; the evidence provided by the imaging studies in combination with the results of recovery following injury have led to the controversial suggestion that language is more unilateral in men than in women.

Sign Language and Bilingualism
Sign language has shown to be lateralized in the left hemisphere of the brain, in the left frontal and temporal lobes. This is known through the use of lesion studies, in which the patients had left hemisphere lesions in the areas associated with language which impaired their ability to sign, while right hemisphere lesions in the same areas show no linguistic deficit (Hickock et al., 1998). Lesions in the right hemisphere of signers did, however, show a limited use of spatial information encoded iconically (which is when the sign is similar-looking to its referent). This is in keeping with the belief that visuo-spatial ability is a right hemisphere function and suggests that the role of the right hemisphere in sign language is in the non-linguistic features of sign language.

Bilingualism is thought to be an overlapping of populations of neurons corresponding to each language, all of which are located in the frontal and temporal regions of the left hemisphere associated with speech comprehension and speech production.

Culture and Language Lateralization
When thinking of language there is a tendency to focus on that language in which you think, however it has been proposed that lateralization of language functions can vary from culture to culture. Asian languages show more bilateral activation during speech than European languages, likely because Asian languages employ a far greater use of right hemisphere abilities, for example prosody, and the use of spatial processing for the more “pictorial” Chinese characters; Native American languages also show a good deal of bilateral activity.

Reorganization following brain injury
Studies have been done following brain injury to determine the level of recovery of language and speech ability, and whether or not recovery is based on lateralized function. Bryan Woods and Hans-Leukas Teuber looked at patients with prenatal and early postnatal brain injury located in either the right or left hemisphere and drew several conclusions. First, if the injury occurs very early, language ability may survive even after left hemisphere brain damage. Second, they found that an appropriation of language regions by the right hemisphere is responsible for the survival of these abilities, but because of this there is a tendency for visuo-spatial ability to be diminished. Third, right hemisphere lesions have the same effect in prenatal and early postnatal patients as they do in adults. Brenda Milner and Ted Rasmussen used the Wada test to determine that early brain injury can cause either left, right or bilateral speech dominance, and that those who retained left hemisphere dominance had damage that was not in either the anterior (Broca’s) or posterior (Wernicke’s) speech zone. Those whose dominance shifted to the right hemisphere most often had damage to these areas. Milner and Rasmussen also found that brain damage which occurs after the age of 5 does not cause a shift in lateralization but rather reorganizes within the hemisphere, potentially employing surrounding areas to take responsibility for some aspects of speech.

In patients who have had hemispherectomy of the left hemisphere, the right hemisphere can often gain considerable language ability. When performed in adulthood, speech comprehension is usually retained (though speech production suffers severe deficits); reading capability is small, and there is usually no writing capability at all.

Learning Exercise: 8 Questions on Hemispheric Language Lateralization
1. In terms of hemispheric lateralization and split-brain patients (individuals which have had commissurotomies), if the word “pencil” was presented to the right field of vision of a split-brain patient and he/she was asked to report what they had seen, the patient would respond:

 * a) by selecting a pencil with the contralateral hand

 * b) by saying the word “pencil”

 * c) by saying “nothing is there”

 * d) by selecting a pencil with the ipsilateral hand

2. The left hemisphere is responsible for all aspects of syntax, except parsing. True or false?

3. What is the structural evidence given to explain the fact that women tend to score higher than men on language-related tasks? What implications might this have on gender differences in patients with aphasia?

4. What 3 conclusions did Bryan Woods and Hans-Leukas Teuber draw regarding the reorganization of language ability following brain injury? Would there be differences in such reorganization in people who are hearing impaired?

5. Through what anatomical system is the right hemisphere able to understand language? What happens to language ability following a removal of the right hemisphere? In what ways do individuals who have had their right hemisphere removed differ from split-brain patients?

6. What were the symptoms of the patient “Tan” which, when presented to neurologist Paul Broca in 1861, propelled Broca to his theory regarding hemispheric language lateralization? Based on current methods of assessment, would Broca's theory still be considered valid today? Why or why not?

7. Which type of study would be best used in order to assess anatomical asymmetry and why?

8. Which type of study is most useful in assessing the connection between hemispheric language lateralization and handedness, and why?


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User talk:PeterSymonds

Your oversight of Request custodian action
Peter, removing the section requesting oversight, you also removed non-sensitive information regarding Moulton behavior, mentioned and referenced in the meta request for locking, and, as well, notice of the meta request. That's creating more work. Would you consider restoring what is not sensitive there, I can't even read it now to check. (There has been plenty of discussion of the outing, that it exists isn't something that could be concealed, overall.) Thanks. --Abd 18:00, 2 March 2011 (UTC)
 * Yes, but technically, the content could be viewed in future diffs, so all the diffs I suppressed contained the same private information. So, no, unfortunately I cannot restore any of it. I appreciate that it may take a little more work but the privacy policy trumps this. Best, PeterSymonds 18:09, 2 March 2011 (UTC)
 * But I can provide the non-sensitive content here:

In addition to the two I link above, and the rather obvious post Moulton has made immediately above, please also remove this. Moulton, refer to me as KC or KillerChihuahua if you expect a response. KillerChihuahua 22:07, 1 March 2011 (UTC)

 * You have made it abundantly clear that you have no intention of addressing my grievance with respect to your wretched conduct here on WV and previously on the English Wikipedia. —Moulton 19:40, 2 March 2011 (UTC)

 * Notice of meta request for global lock. See . This request, if accepted by a steward, would return the situation to the status quo prior to the Ottava unblock of Caprice. It would still allow local control, though any local custodian acting, perhaps to whitelist-exempt one or both of these accounts, should certainly consider what is necessary for protection of users from outing and harassment. --Abd 17:27, 2 March 2011 (UTC)

 * I regret that I have but a quarter-million lives to give to my community. —Caprice 17:34, 2 March 2011 (UTC)

Thanks, Peter, that should help. I'll replace that on RCA if you have not done so.
 * You're welcome (I didn't add it back; I'd rather leave it to the locals). Feel free to e-mail myself or the stewards list if there are further issues. PeterSymonds 18:24, 2 March 2011 (UTC)

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Talk:The resolution of the problem of the Prime Sequence

The resolution of the problem of the Prime Sequence. [THE RESOLUTION OF THE PRIME SEQUENCE PROBLEM] The reader who is interested in this problem - an unsolved mystery - of the erratic prime sequence is respectfully redirected to the contribution having the shorter title: " The Prime Sequence Problem " where ordered groups of three primes and another uneven integer, their negative summation, are treated as the four co-ordinates of points in a four dimensional Euclidean space.

This latter contribution is in the process of being written and is at the stage of requiring a knowledge of [well established] linear algebra to progress the matter.

A discrete version of the Legendre Transformation is a key concept in the development of the subject.

There is of course much other literature on the above subject of the prime sequence, but as far as is known, the treatment in the above shorter titled contribution is uniquely original.SHAWWPG19410425 (talk) 23:31, 1 December 2012 (UTC)


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User:Eml5526.s11.team3.sahin/Homework 47

=Problem 5.3 Using Lagrange Interpolation Basis Functions to Obtain Trial and Exact Solution of Torsion Bar=

Given: Data set G1DM1.0/D1b defined on Lecture 26-2
Strong form for the Torsion Bar

where $${a_2}\left( x \right){\text{ =  2 + 3x}}$$  $$\qquad\text{and}\quad $$   $$f\left( {x,t} \right) = 5x $$  so we get Essential boundary conditions, Natural boundary conditions,

Find

1. Explain how Langrange Interpolation Basis Functions (LIBF) are used as Constraint Breaking Solutions

2. Plot all LIBF

3. Use matlab quad, WolframAlpha, ... to integrate

4. Plot $$ u_m^h  \ vs\   u$$ and $$ u_m^h\left(0.5 \right)-u\left(0.5 \right)\ vs\ m$$

1. Explaining how LIBF can be used as CBS
For a basis function, and for $$\displaystyle{{\Gamma }_{g}}=\{\beta \}$$, the corresponding basis functions satisfying

{ $$\displaystyle {{b}_{j}}(x), \quad j=1,2...n $$ }   such that   $$\displaystyle {{b}_{1}}(\beta)\ne 0$$   and     $$\displaystyle {{b}_{j}}(\beta)= 0, \quad  j = 2,3...n $$

Lets show that the statement above can be satisfied to explain how LIBF can be used as CBS

We know that general formula of Lagrange Interpolation Basis Function is

Lets denote m=4,

If $$\displaystyle x={{x}_{1}}, \qquad {b}_{1}=L_{1,4}(x_{1})= \delta_{1,1} =\frac{(x_{1}-x_{2})(x_{1}-x_{3})(x_{1}-x_{4})}{(x_{1}-x_{2})(x_{1}-x_{3})(x_{1}-x_{4})} =1 $$ If $$\displaystyle x={{x}_{2}},\qquad {b}_{2}=L_{1,4}(x_{2})= \delta_{1,2}= \frac{(x_{2}-x_{2})(x_{2}-x_{3})(x_{2}-x_{4})}{(x_{1}-x_{2})(x_{1}-x_{3})(x_{1}-x_{4})} = 0$$

If $$\displaystyle x={{x}_{3}},\qquad {b}_{3}= L_{1,4}(x_{3})= \delta_{1,3}= \frac{(x_{3}-x_{2})(x_{3}-x_{3})(x_{3}-x_{4})}{(x_{1}-x_{2})(x_{1}-x_{3})(x_{1}-x_{4})} = 0$$

If $$\displaystyle x={{x}_{4}}, \qquad {b}_{4}= L_{1,4}(x_{4})= \delta_{1,4}=\frac{(x_{4}-x_{2})(x_{4}-x_{3})(x_{4}-x_{4})}{(x_{1}-x_{2})(x_{1}-x_{3})(x_{1}-x_{4})} = 0$$

So for the Lagrange basis function $$\displaystyle {{b}_{j}}(x) $$ satisfy the Constraint Breaking Solution requirements since $$\displaystyle {b}_{1} =1, \quad and \quad {b}_{2}={b}_{3}={b}_{4} = 0 $$

And we can conclude that

We do same calculation for other m values such as m = 2,3,5,6,7,8,9 in Matlab program. As a result, we can choose Lagrange interpolation function as a basis because it will destroy the constraint.

2.Plotting all LIBF used for m = from 2 to 9
The LIBF used must satisfy the essential boundary conditions $$\displaystyle u\left( x=0 \right) = 4 \quad $$  according to Discrete Weak Form.

The essential boundary conditions to be satisfied are

$$\displaystyle {{b}_{1}}(0)=1 $$ and  $$\displaystyle {{b}_{2...n}}(0)=0 $$

Hence to satisfy the condition, the LIBF basis functions used

Using these basis functions, the LIBF are plotted for m = 2,3,4,5,6,7,8, and 9 until the convergence of the basis functions with the actual solution takes place where the error is less than $$\displaystyle  {{10}^{-6}} $$

The convergence becomes less than $$\displaystyle  {{10}^{-6}} $$ for  $$\displaystyle  m=9 $$. It is 1.0266 e-07 The LIBF are plotted for various values of $$\displaystyle m $$ until convergence.

The plots of the basis functions were developed using Matlab. These LIBF plots were produced by the following Matlab code below.

i. Constracting of Exact Solution
We have general 1-Dimensional Model Data set 1.b from Eq 3.1 to Eq 3.5

Integrating the strong form Eq 3.2 with respect to x, we get,  -

From the natural boundary condition,

If we denote natural boundary condition, we obtain

If we substitute $$ \displaystyle C_{1} $$ into Eq 3.11

Integrating Eq 3.15 again with respect to x using integration by parts with constant $$ \displaystyle C_{2}$$ involved, we have

We already know this from Homework 2.4

If we take it out of integral, we have

Now using essential boundary condition with $$ \displaystyle u \left( {x = 0} \right) =4$$ ,we can update above equation as -

So we find

Substituting this constant values gives

It is also equal to

ii. Constructing of Weak Form
As the weight function must vanish on the essential boundaries, we consider all smooth weight functions $$w(x)$$ such that $$w(0)= 0$$.

Multiplying the given differential equation and the natural boundary condition over the domain specified, by an arbitrary weight function

Now we integrate the Equation 3.24 by parts,

Integration by parts:

So

w(0) =0 therefore, the first term on the RHS of the above vanishes at x = 0. Substituting Equation 3.27 in Equation 3.24,

Substituting Equation 3.25 in Equation 3.28,

iii. Discretization of Weak Form
Approximated solution $$ u^{h} \left ( x \right ) $$ and $$ w^{h} \left ( x \right ) $$ :

where $$ \left ( c_{i} \right ) $$ and $$ \left ( d_{j} \right ) $$ are constants and $$ \left ( b_{i} \right ) $$ and $$ \left (b_{j}\right ) $$ are the LIBF. We divide equally to discretizate on our domain for m=2,3,4,5,6,7,8,9. We will use these LIBF as a shape function satisfying the CBS.

When we expand (Eq 3.30) for = 4, we get

where,

We get matrices K and F calculated in Matlab code below.

The convergence occurs at $$\displaystyle m=9 $$

Using Matlab to construct above two matrices until satisfying the convergence criterion, we finally have
 * {| style="width:100%" border="0"

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

{{K}_{FF}}=\left( \begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\   -48.861 & 167.423 & -276.316 &  345.621 & -349.434 & 260.149 & -142.253 & 55.498 & -11.827 \\    58.104 & -276.316 & 593.033 & -828.416 & 880.332 & -718.901 & 440.528 & -193.210 & 44.845 \\   -74.821 & 345.622 & -828.416 & 1349.981 & -1583.123 & 1382.496 & -919.426 & 432.555 & -104.867 \\    72.070 & -349.434 & 880.332 & -1583.123 & 2080.253 & -1954.709 & 1351.427 & -660.485 & 163.668\\    -49.752 & 260.150 & -718.901 & 1382.497 & -1954.709 & 2000.226 & -1453.569 & 712.071 & -178.013 \\    24.315 & -142.253 & 440.528 & -919.426 & 1351.427 & -1453.569 & 1152.701 & -595.975 &  142.252 \\    -8.185 &  55.498 & -193.211 &  432.555 & -660.485 &  712.071 & -595.975 & 376.137 & -118.406\\    1.520 & -11.827 & 44.845 &  -104.867 & 163.668 & -178.013 &  142.252 & -118.407 &  60.828\\ \end{matrix} \right)

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

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

{{F}_{F*1}}=\left( \begin{matrix}  4\\   0.130\\  -0.041\\    0.694\\  -0.400\\   1.157\\ -0.123\\   0.909\\   12.174\\ \end{matrix} \right)

$$
 * }

By the equation $$\displaystyle Kd =F $$, we have
 * {| style="width:100%" border="0"

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

{{d}_{F*1}}=\left( \begin{matrix}  4.0000\\    4.8299\\    5.5341\\    6.1415\\    6.6712\\    7.1363\\    7.5463\\    7.9085\\    8.2283\\ \end{matrix} \right)

$$
 * }


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title: File:FBDig.png


title: File:Prob6.jpg


title: File:Nm1.s11.mtg26.djvu


text:
Digital Age/Moore's Law

Technology is changing rapidly. Just how rapidly is fascinating - in many areas of technology we're seeing exponential growth, leading to change that is occurring faster than we may be able to follow. Moore's Law points to the extent of this change, so in this module we're going to start with Moore's Law and expand beyond, looking at what this level of change may mean for the future. This will include looking at the notion of the technological singularity, and we'll do a quick examination of some of the drivers and enablers of this change.

Moore's Law
The Intel Corporation, today known for their work in microprocessor technology, were founded in 1968 by Robert Noyce and Gordon Moore. Among the company's many achievements, Intel developed the first microprocessor in 1971, and went on to develop the "x86" line of microprocessors - including the Pentium CPUs found in many modern microcomputers.

Prior to founding Intel, in 1968, Gordon Moore proposed what is now known as "Moore's Law", often expressed as: "The number of transistors on an integrated circuit will double every two years". (Strictly speaking, he originally said every year, but it was changed to every two years in 1975). It is an interesting statement - what it means, in effect, is that every two years the power of computers will double, noting that "number of transistors" doesn't necessarily equate to "computational power".

Moore proved to be, on the whole, correct. Approximately every two years the complexity of integrated circuits has doubled, while their price has continued to fall. This is an impressive achievement - if, for example, cars had followed the same scale, we would have gone from a top speed of 16 kph with the Patent-Motorwagen in 1865, to a top speed of 32,768 kph in 1906. Nevertheless, similar laws exist for other technologies:

 * Kryder's Law: Magnetic disk storage density doubles every year.
 * Nielsen's Law: The bandwidth available to home users will increase by 50% each year.
 * Hendy's Law: The number of pixels per dollar in a digital camera doubles every two years.

Through your own experiences you'll have encountered many of these - if you try buying a hard drive today, compare what you could buy for the same price as what you would have been able to purchase twelve months ago, or look at the pixel count on a entry-level digital camera. The different speeds of development are also interesting: combining Hendry's Law with Kryder's Law, we can be confident that we can buy sufficient storage to meet out photographic needs for the next few years, but our ability to share those pictures online might be hindered by the rate at which network speeds are increasing.

Law of Accelerating Returns
Although there are other laws similar to Moore's Law, Moore's Law itself only applies to a limited range of technology, and even then there would appear to be an upper limit as to how long it can continue. Eventually we'll hit physical limits as to how many transistors can be fit onto a microprocessor, although the point at which this will occur is unclear, some current theories suggest that it will be reached in less than ten years. Similarly, if we want to look at this long term, it has been argued that there are limits to how much information can be contained in the universe, and it is suggested that Moore's Law will hit that limit in 600 years (Kraus and Starkman, 2004).

On the plus side, at least some of these limitations can be overcome if we include new technologies in the picture. Raymond Kurzweil is one of those who have argued along these lines, arguing that once the limits of a particular type of technology are reached, a paradigm shift will see a new type of technology replacing the old, continuing the growth. If this is true, Kurzweil suggests that in 2045 the rate of change in technology will bring about a technological singularity, where the rate of change is so fast it will "rupture" human history.

Drivers and enablers of change
Whether or not you accept that this exponential growth will continue, one thing is clear: things are changing fast. Technology, and the potential impact that it can have, is moving forward rapidly, and that rate of change is likely to continue to increase. Thus if you find it uncomfortable facing the way the world is changing now things aren't going to get better. On the other hand, the times will certainly be interesting.

But be that as it may, change doesn't always happen for its own sake. In this case, I'm going to present some possible drivers for change, and, for the sake of discussion, I'm separating them into two categories: technological enablers that make the change possible and social drivers that create reasons for the change. Significantly, these don't always go together - just because there's a desire for something doesn't mean that the technology exists to make it happen, and just because technology can produce something, we shouldn't assume that there is a desire to see that produced.

Processors/Performance
Obviously, more powerful computers can do more things – or so you might think. The Church–Turing thesis effectively states that any program that can be completed in a finite time can be completed by the simplest computer thus proposed – the Turing Machine. In short, given certain limitations (memory and speed) any Turing-complete computer can perform any computing task that can be completed.

Mind you, those limitations are important: there's not much point in a computer that takes so long to complete a task that we won't see the results in our lifetime (unless the computer concerned is Deep Thought, and the task is calculating the ultimate question to life, the universe and everything). Similarly, if the computer doesn't have enough memory, it might run out before completing the task: the Turing Machine's solution was to use an infinitely long tape, but they can be hard to come by.

Networks
Networks give us the ability to share data, and this is core to our ability to communicate, whether through personal interaction, telephones or computers. Increases in network speed improves they type of data that can be shared and the speed at which we can share it, while networks also help address the question of storage: we don't need to store everything on the computer, but can instead store it online and download as required. If you are interested in this side of things, you may wish to look into cloud storage.

Size
Size is also a key enabler: large devices are cool, but smaller devices are more portable, very small devices open up a range of possibilities. Going right down to nanotechnology, devices on a microscopic scale open up a wealth of possibilities that we still don't fully comprehend.

Cost
Finally, there isn't much point having exciting new technology if no-one can afford it. Cost reductions take technology from specific, high end tasks, and open up the use of that technology in far more mundane areas. I've often been taken by the way we "waste" technology - if we took an iPad back to 1950 and showed them what it could do, the computer scientists of the day would be astounded. But they would be even more astounded when we show them how all that computational power, of a type that they may not have even dreamed of, is being used to play Angry Birds.

Communication
From the phone, through email, and on to video chat – with lots in between. We're using more advanced technologies not just to communicate in different ways, but also to increase the level and frequency of our communications. If we went a few years back, who would have expected Twitter to take off - a system whereby you share with your "followers" every small aspect of your life in almost real time. In terms of enablers, the key one is probably a growth in networks, but cost, size and performance all play a role.

Desire for collaboration
Collaboration needs tools that facilitate it, and technology is helping to develop those tools. These range from improved methods of communication, through to shared spaces such as Wikipedia and, to use a self-referential example, Wikiversity. The big enablers, of course, are networks.

Need to collect and access information
It’s handy having the ability to collect and access information when we need it. This ranges from what we see as traditional "database-type" information, such as who is enrolled in a course or what the average age of Twilight fans is, to material that we traditionally gained from other sources - such as how to get to a friend's house to information about medical problems. For some of these, especially the database style questions, performance and storage are key enablers. But other issues, networks, size and cost come into play.

User interfaces
How we interact with tools to get information or to communicate is another key issue. Different methods of interacting with technology opens up different possibilities as to how it can be used. Indeed, as will be shown when we talk about TAM, the ability to use technology – and, most importantly, the perceived ease by which the technology can be employed – is a major aspect to whether or not the technology will be accepted.

Visualization
The ability to visualize data (whether it be an object or just information) requires hardware and software that can potentially work with massive volumes of data.

Entertainment
I tend to think that we downplay the value of entertainment in pushing technology. If you have broadband at home, why did you purchase it? Was it to improve your ability to work from home, or was it download music, amuse yourself online and play games? Entertainment pushes for and requires faster networks, faster processors and better storage, just as a start.

Research questions

 * Do you believe that Moore's Law will continued to hold?
 * Moore once stated that "Moore's law is a violation of Murphy's law. Everything gets better and better". Is there any truth in this?
 * To what extent do you think technology is developed for entertainment, as opposed to "real work"? What other pursuits are major drivers for technological change?

title: Digital Age/Moore's Law


title: File:Fe1.s11.mtg26.djvu


text:
User:Eml5526.s11.team7/HW4

Problem Description
Re do Problem 2.9, fixing $$\mathbf{\phi=\pi/2}$$ $$\Rightarrow \mathbf{F} = \left \{ 1, sinx, sin2x, sin3x, \cdot \cdot\cdot\cdot \right \}$$

For n=2
PART 1 From G1D1M.1.0/D2 $$P(u)=0 = \frac{\partial }{\partial x}\left[ 2\frac{\partial u}{\partial x}\right] + 3$$ therefore the exact solution is $$u(x)=\frac{-3}{4}x^2-4x+\frac{19}{4} $$ $$ d=\left \{ d_0, d_1, d_2 \right \}$$ $$ b=\left \{ 1,sinx,sin2x \right \}$$ $$ u^h(x) = \sum_{j=0}^2{d_jb_j(x)}=d_0 + d_{2}sin(2x) + d_{1}sin(x)$$ $$ (u^h)' =\frac{d{u}^{h}}{dx}(x) = \sum_{j=0}^2{d_jb_j'(x)}=d_{1}cos(x) + 2d_{2}cos(2x)$$ PART 2 The boundary conditions are enforced to generate two equations as follows: $$\displaystyle u^h(x=1) =0=d_{0} + 0.8414d_{1} + 0.9092d_{2}$$ $$\displaystyle (u^h)'(x=0) =-4=d_{1} + 2d_{2}$$  PART 3  An additional equation is generated by projected the residue onto a basis function as follows: $$\int_{0}^{1}b_{k}(x)P(u^h)dx=-0.9193d_1 - 5.664d_2=-3$$ PART 4  The equations are assembled into stiffness coefficient and force matrices in the form $$\mathbf{Kd}=\mathbf{f}$$:

$$ \begin{bmatrix} 1.0000 &   0.8415 &   0.9093\\         0   & 1.0000  &  2.0000\\         0  & -0.9194  & -5.6646   \end{bmatrix} \begin{bmatrix} d_0\\ d_1\\ d_2 \end{bmatrix} = \begin{bmatrix} 0\\ -4\\ -3 \end{bmatrix} $$ Note that the stiffness matrix, $$\mathbf{K}$$, is not symmetric PART 5  The coefficients are generated by solving the matrices above yielding:

$$ d=\begin{pmatrix} 4.7162\\  -7.4908\\    1.7454 \end{pmatrix} $$

For n=4
$$ d=\begin{pmatrix} 4.7491\\ -13.1144\\    8.2087\\   -3.6116\\    0.8829 \end{pmatrix} $$

For n=6
$$ d=\begin{pmatrix} 4.7500\\ -257.2135\\ 341.5393\\ -261.6484\\  128.7013\\  -38.5824\\    5.5312 \end{pmatrix} $$

Author
Eml5526.s11.team7.Srilalithkumar 05:05, 2 March 2011 (UTC)

Problem Description
“A First Course in Finite Elements” by Fish & Belytschko P73 Problem 3.5

Obtain the weak form of the equations of heat conduction with the following boundary conditions: The essential boundary condition defined at $$x=0$$ $$T(0) = 100$$ and the natural boundary condition defined at $$x=10$$ $$\bar q(10) = hT$$. The natural boundary condition is a convection condition.

Solution
The strong form of the heat problem is:
 * {| style="width:100%" border="0"

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

\frac{d}(AK\frac) + s = 0

$$     (4.2.1)
 *
 * }
 * {| style="width:100%" border="0"

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

T(0) = 100

$$     (4.2.2)
 *
 * }
 * {| style="width:100%" border="0"

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

\bar q(10) = hT

$$     (4.2.3) So the weak form is obtained as follows:
 *
 * }
 * {| style="width:100%" border="0"

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

\int_0^{10} {w(\frac{d}(AK\frac) + s)} dx = 0

$$     (4.2.4)
 *
 * }
 * {| style="width:100%" border="0"

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

\Rightarrow wAK\frac|_0^{10} - \int_0^{10} {\fracAK\fracdx + \int_0^{10} {wsdx} } = 0

$$     (4.2.5)
 *
 * }
 * {| style="width:100%" border="0"

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

\Rightarrow w(10)AK\frac - w(0)AK\frac - \int_0^{10} {\fracAK\fracdx + \int_0^{10} {wsdx} } = 0

$$     (4.2.6)
 *
 * }
 * {| style="width:100%" border="0"

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

\Rightarrow w(10)AhT - 0 - \int_0^{10} {\fracAK\fracdx + \int_0^{10} {wsdx} } = 0\;\;at\;\;w(0) = 0

$$     (4.2.7)
 *
 * }
 * {| style="width:100%" border="0"

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

\Rightarrow w(10)AhT + \int_0^{10} {wsdx} = \int_0^{10} {\fracAK\fracdx}\;\;\;\; \forall w\;|\;w(0) = 0

$$     (4.2.8 )
 *
 * }

To sum up, the weak form of the heat problem is:
 * {| style="width:100%" border="0"

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

w(10)AhT + \int_0^{10} {wsdx} = \int_0^{10} {\fracAK\fracdx}\;\;\;\; \forall w\;|\;w(0) = 0

$$     (4.2.9)
 *
 * }
 * {| style="width:100%" border="0"

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

T(0) = 100

$$     (4.2.10)
 *

 * }

Author
Zongyi Yang

Brandon Hua

Problem Description

 * [From Fish & Belytschko, Chapter 3, Problem 3.6]

Given the strong form for the heat conduction problem in a circular plate:

 * $$k \frac{d}{dr}\left( r \frac{dT}{dr}\right)+rs=0 \qquad \text{on} \qquad 0 < r \leq R.$$

 * natural boundary condition: $$\displaystyle \frac{dT}{dr}(r=0) = 0,$$

 * essential boundary condition: $$\displaystyle T(r=R) = 0,$$

 * where R is the total radius of the plate, s is the heat source per unit length along the plate radius, T is the temperature and k is the conductivity. Assume that k, s, and R are given.

Find

 * 1) Construct the weak form for the above strong form.
 * 2) Use quadratic trial (candidate) solutions of the form $$\displaystyle T = \alpha_0 + \alpha_1 r + \alpha_2 r^2$$ and weight functions of the same form to obtain a solution of the weak form.
 * 3) Solve the differential equation with the boundary conditions and show that the temperature distribution along the radius is given by
 * $$ \displaystyle T = \frac{s}{4k} (R^2 - r^2)$$.

Part 1
First, the governing equation and natural boundary condition are multiplied by the weight function, $$\displaystyle w(r) $$, and integrated over the domain ($$ 0 < r \leq R$$). The restriction on the weight function is that $$\displaystyle w(R)=0$$.

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

$$ \int_{0}^{R}w(r) \left [ k \frac{d}{dr} \left ( r \frac{dT}{dr} \right ) + rs \right ] dr = 0 \qquad \text{and} \qquad w(r) \frac{dT}{dr} = 0.$$ (4.3.1)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

The first equation can be expanded as

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

$$ \int_{0}^{R}w(r) k \frac{d}{dr} \left ( r \frac{dT}{dr} \right ) dr + \int_{0}^{R}w(r) r \ s \ dr = 0 \qquad \forall w.$$ (4.3.2)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

The following integration by parts is simplified if the substitution $$\displaystyle f = r \frac{dT}{dr}$$ is made. Also note that the weight function will subsequently be referred to as $$\displaystyle w$$ with the dependence on the spatial domain, $$\displaystyle r $$, being implied. Equation 4.3.2 can therefore be written as,

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

$$ \int_{0}^{R}w k \frac{df}{dr} dr + \int_{0}^{R}wrs \ dr = 0.$$ (4.3.3)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

Integration by parts yields,

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

$$ \int_{0}^{R} \frac{d}{dr} (wkf)dr - \int_{0}^{R} fk \frac{dw}{dr} dr + \int_{0}^{R}wrs \ dr = 0$$ (4.3.4)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

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

$$ wkf|_{0}^{R} - \int_{0}^{R} kf \frac{dw}{dr} dr + \int_{0}^{R}wrs \ dr = 0.$$ (4.3.5)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

The substitution in equation 4.3.3, $$\displaystyle f = r \frac{dT}{dr}$$, can now be reintroduced, and equation 4.3.5 becomes

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

$$ \left( wkr \frac{dT}{dr} \right)_{r=R} - \left( wkr \frac{dT}{dr} \right)_{r=0} - \int_{0}^{R} kr \frac {dT}{dr} \frac{dw}{dr} dr + \int_{0}^{R}wrs \ dr = 0.$$ (4.3.6)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

The first term reduces to zero due to the prescribed weight function at $$\displaystyle r=R $$. In other words, $$\displaystyle w(R)=0 $$. The second term also reduces to zero since the natural boundary condition defines the heat flux at $$\displaystyle r=0 $$ to equal zero. Therefore, the weak form of the heat conduction problem becomes

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

$$ \int_{0}^{R} \frac{dw}{dr} kr \frac {dT}{dr} dr = \int_{0}^{R}wrs \ dr \qquad \forall w \quad \text{with} \quad w(R)=0.$$ (4.3.7)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Part 2
Consider the candidate and weight solutions of the following forms:

 * Candidate: $$\displaystyle T(r) = \alpha_0 + \alpha_1 r + \alpha_2 r^2$$,
 * Weight: $$\displaystyle w(r) = \beta_0 +\beta_1 r + \beta_2 r^2 $$.

For the candidate solution to be admissible, it must satisfy the essential boundary condition T(R) = 0. Likewise, the weight function must vanish at r = R. Implementing these conditions yields expressions with respect to the current and total radii of the plate.

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

$$ \displaystyle T(R) = \alpha_0 + \alpha_1 R + \alpha_2 R^2 = 0 \qquad \Rightarrow \qquad \alpha_0 = - \alpha_1 R - \alpha_2 R^2 $$     (4.3.8)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

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

$$ \displaystyle w(R) = \beta_0 +\beta_1 R + \beta_2 R^2 = 0 \qquad \Rightarrow \qquad \beta_0 = - \beta_1 R - \beta_2 R^2 $$     (4.3.9)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

Substituting these expressions for $$\displaystyle \alpha_0$$ and $$\displaystyle \beta_0$$ back into the candidate solution and weight function,

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

$$ \displaystyle T(r) = \alpha_1 \left( r-R \right) + \alpha_2 \left( r^2 - R^2 \right) $$     (4.3.10)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

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

$$ \displaystyle w(r) = \beta_1 \left( r-R \right) + \beta_2 \left( r^2 - R^2 \right) $$     (4.3.11)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

The derivatives of equations 4.3.10 and 4.3.11

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

$$ \frac{dT}{dr} = \alpha_1 + 2 \alpha_2 r,$$ (4.3.12)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

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

$$ \frac{dw}{dr} = \beta_1 + 2 \beta_2 r. $$ (4.3.13)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

Equations 4.3.10 through 4.3.13 are substituted into the weak form, equation 4.3.7:

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

$$ \int_{0}^{R} \left( \beta_1 + 2 \beta_2 r \right) kr \left( \alpha_1 + 2 \alpha_2 r \right) dr = \int_{0}^{R} \left[ \beta_1 (r-R) + \beta_2 (r^2 - R^2) \right] rs \ dr $$ (4.3.14)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

Integrating the both sides of the equation over the domain yields

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

$$ \text{LHS:} \quad \int_{0}^{R} \left( \beta_1 + 2 \beta_2 r \right) kr \left( \alpha_1 + 2 \alpha_2 r \right) dr \quad = \quad \frac{2}{3} \alpha_1 \beta_2 k R^3 \ + \ \frac{1}{2} \alpha_1 \beta_1 k R^2 \ + \ \frac{2}{3} \alpha_2 \beta_1 k R^3 \ + \ \alpha_2 \beta_2 k R^4 $$     (4.3.15)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

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

$$ \text{RHS:} \quad \int_{0}^{R} \left[ \beta_1 (r-R) + \beta_2 (r^2 - R^2) \right] rs \ dr \quad = \quad - \frac{1}{4} \beta_2 s R^4 \ - \ \frac{1}{6} \beta_1 s R^3 $$     (4.3.16)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

Equations 4.3.15 and 4.3.16 are substituted into 4.3.14, and the entire equation is rewritten to collect the terms $$\displaystyle \beta_1$$ and $$\displaystyle \beta_2$$.

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

$$ \beta_1 \left( \frac{1}{2} \alpha_1 k R^2 + \frac{2}{3} \alpha_2 k R^3 + \frac{1}{6} s R^3 \right ) + \beta_2 \left( \frac{2}{3} \alpha_1 k R^3 + \alpha_2 k R^4 + \frac{1}{4} s R^4 \right) =0 $$     (4.3.17)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

The coefficients $$\displaystyle \alpha_1$$ and $$\displaystyle \alpha_2$$ can be determined in terms of the known components s, k, and R.

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

$$ \displaystyle \begin{bmatrix} {\frac{1}{2}R} & { \frac{2}{3}R^3} \\ \\ {\frac{2}{3}R^3} & {R^4} \end{bmatrix} \cdot \begin{bmatrix} \alpha_1\\ \\ \alpha_2 \end{bmatrix} = \begin{bmatrix} {- \frac{1}{6}sR^3} \\ \\ {- \frac{1}{4}sR^4} \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$     (4.3.18)
 *

 * }

 * Therefore,

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

$$ \alpha_1 = - \frac{ks}{12} \left( 2R^7 + R^5 \right) \qquad \text{and} \qquad \alpha_2 = - \frac{ks}{36} \left( 9R^8 + 4R^6 \right) $$     (4.3.19)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

Substituting the values of $$\displaystyle \alpha_1$$ and $$\displaystyle \alpha_2$$ into equation 4.3.10 yields the candidate solution in terms of known values. There is some concern regarding the validity of this equation, since the system units do not correspond to an appropriate unit measurement of temperature.

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

$$ T(r) = - \frac{ks}{12} \left( 2R^7 + R^5 \right)(r-R) - \frac{ks}{36} \left( 9R^8 + 4R^6 \right)(r^2 - R^2) $$     (4.3.20)
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The candidate (trial) solution above is substituted into the governing equation to confirm equilibrium.

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$$ k \frac{d}{dr}\left( r \frac{dT}{dr}\right)+rs=0 \qquad \Rightarrow \qquad -k \left[ \frac{ks(2R^7 + R^5)}{12} + \frac{rks(9R^8 + 4R^6)}{9} \right] +rs = 0 $$     (4.3.21)
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 * Solving for $$\displaystyle r$$ and simplifying, we are given an expression for when $$\displaystyle r$$ is valid for given values of k and R. Note that the validity of $$\displaystyle r$$ is independent of the heat source per unit length, s.

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$$ r = - \frac{6R^7 k^2 + 3R^5k^2}{36R^8k^2 + 16R^6k^2 - 36} $$     (4.3.22)
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Part 3
The differential - or governing - equation below will now be solved for $$\displaystyle T(r)$$.

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$$ k \frac{d}{dr}\left( r \frac{dT}{dr}\right)+rs=0 \qquad \text{on} \qquad 0 < r \leq R $$ (4.3.23)
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The equation is rewritten with $$\displaystyle rs$$ and $$\displaystyle k$$ moved to the right hand side. Then both sides can be integrated over $$\displaystyle r$$.

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$$ \int \frac{d}{dr} \left( r \frac{dT}{dr} \right) dr = - \int \frac{s}{k} r dr $$ (4.3.24)
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This reduces to

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$$ r \frac{dT}{dr} = - \frac{s}{2k} r^2 + constant $$     (4.3.25)
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Recalling the natural boundary condition and applying it to equation 4.3.25,

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$$ \frac{dT}{dr}(0) = 0 \quad \Rightarrow \quad r \frac{dT}{dr} = - \frac{s}{2k} r^2 + constant $$     (4.3.26)
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$$ 0 = 0 + constant \qquad \text{therefore} \qquad r \frac{dT}{dr}  = - \frac{s}{2k} r^2. $$     (4.3.27)
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Rewrite as

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$$ dT = - \frac{s}{2k} r dr \qquad \Rightarrow \qquad \int dT  = - \int \frac{s}{2k} r dr. $$ (4.3.28)
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Integrate recalling the essential boundary condition, $$\displaystyle T(R) = 0$$.

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$$ T = - \frac{s}{4k} r^2 + constant \quad \Rightarrow \quad T(R) = - \frac{s}{4k} R^2 + constant = 0 $$     (4.3.29)
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 * $$\quad \Rightarrow constant = \frac{s}{4k} R^2$$

This confirms that the temperature distribution along the radius is given by

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$$ T(r) = \frac{s}{4k} \left( R^2 - r^2 \right). $$     (4.3.30)
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Authors
Philip Flater Braden Snook

Problem Description
HW 4.4 p.21.1 Problem 3.7 from "A First Course in Finite Elements" by Fish and Belytschko

Given the strong form for the circular bar in torsion:
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$$ \displaystyle \frac{d}{dx}\left ( JG\frac{d\phi }{dx} \right )+m=0, 0\leq x\leq l $$ (4.4.1)
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Essential Boundary Condition:
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$$ \displaystyle \phi \left ( x=0 \right )=\overline{\phi } $$       (4.4.2)
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Natural Boundary Condition:
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$$ \displaystyle M\left ( x=l \right )=\left ( JG\frac{d\phi }{dx} \right )_{l}=\overline{M} $$       (4.4.3)
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Where:

$$ \displaystyle m(x) $$ is a distributed moment per unit length $$ \displaystyle M $$ is the torsion moment $$ \displaystyle \phi $$ is the angle of rotation $$ \displaystyle G $$ is the shear modulus $$ \displaystyle C $$ is the radius of the circular shaft $$ \displaystyle J $$ is the polar moment of inertia given by the equation

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$$ \displaystyle J=\frac{\pi C^{4}}{2}$$ (4.4.4)
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a) Contruct the weak form for the circular bar in torsion. b) Assume that (m=0) and integrate the differential equation given above (4.4.1). Find the integration constants using the boundary conditions (4.4.2) and (4.4.3).

Hint: "I suggest to use this opportunity to solve the G1DM1.0 / D2 (at least a slightly modified version of it) as stated on p.9-2 of Mtg 9 (e); you just need to swap the boundary locations around to conform to the statement for HW 4.4, i.e., let x=0 be the location for the ess. b.c., and x=1 the location for the nat. b.c. So use the same data as stated in G1DM1.0 / D2, with the boundary locations swapped around; we can call this problem G1DM1.0 / D2b (data set 2b). This way, you can plot the exact solution, which can then be used to verify the results for future problems." -Prof Vu Quoc

Part a
Mulitply (4.4.1) by an arbitrary function $$ \displaystyle w(x) $$ and integrate over the domain $$ \displaystyle \Omega = ]0,l[ $$

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$$ \displaystyle \int_{0}^{l}w\left [ \frac{d}{dx}JG\frac{d\phi}{dx} \right ]dx+\int_{0}^{l}wMdx=0, \forall w $$ (4.4.5)
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$$ \displaystyle \frac{d}{dx}\left ( wf \right )=\frac{d(wf)}{dx}-f\frac{dw}{dx} $$       (4.4.6)
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$$ \displaystyle \int_{0}^{l}w\frac{dw}{dx}=\int_{0}^{l}\frac{d}{dx}(wf)dx-\int_{0}^{l}f\frac{dw}{dx}dx $$       (4.4.7)
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$$ \displaystyle \int_{0}^{l}w\frac{dw}{dx}=(wf)_{x=l}-(wf)_{x=0}-\int_{0}^{l}f\frac{dw}{dx}dx $$       (4.4.8)
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Let
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$$ \displaystyle f=JG\frac{d\phi}{dx} $$       (4.4.9)
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(4.4.5) becomes
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$$ \displaystyle \left (wJG\frac{d\phi}{dx} \right )_{x=l}-\left (wJG\frac{d\phi}{dx}  \right )_{x=0}-\int_{0}^{l}\frac{dw}{dx}JG\frac{d\phi}{dx}dx-\int_{0}^{l}wmdx=0,\forall w, w(l)=0 $$       (4.4.10)
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$$ \displaystyle G\frac{d\phi}{dx}=\tau $$       (4.4.11)
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(4.4.10) becomes
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$$ \displaystyle \left (wJ \tau  \right )_{x=l}-\left (wJ \tau   \right )_{x=0}-\int_{0}^{l}\frac{dw}{dx}JG\frac{d\phi}{dx}dx-\int_{0}^{l}wmdx=0,\forall w $$ (4.4.10)
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Recall:

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$$ \displaystyle w(0)=0 $$       (4.4.12)
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Hint: "The problem stated on p.9-2 of Mtg 9 (e) corresponds to the G1DM1.0 / D1 (data set 1, in which the coefficient a_2 is a linear function of x), and NOT the G1DM1.0 / D2 (data set 2, in which the coefficient a_2 is a constant). To be precise, the problem that I referred to on p.9-2 of Mtg 9 (e) is stated in Eq.(3) on that page.  Hence, let's define G1DM1.0 / D1b (data set 1b, NOT 2b) as the problem when you swap the boundary conditions around, i.e., let x=0 be the location for the ess. b.c., and x=1 the location for the nat. b.c." -Prof Vu-Quoc From (3) 9-2

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$$ \displaystyle \frac{d}{dx}\left [ \left ( 2+3x \right ) \frac{du}{dx} \right ]+5x=0,\forall x\varepsilon ]0,1[ $$ $$ \displaystyle \Rightarrow JG=2+3x $$ $$ \displaystyle \Rightarrow m(x)=5x $$       (4.4.13)
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$$ \displaystyle u(0)=4 $$ $$ \displaystyle \Rightarrow \phi(0)=4 $$       (4.4.14)
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$$ \displaystyle \frac{du}{dx}(x=1)=-6  $$ $$ \displaystyle \Rightarrow \frac{d\phi}{dx}(x=1)=-6 $$       (4.4.15)
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Let
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$$ \displaystyle w(0)=0 $$ (4.4.10) becomes:
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$$ \displaystyle \left (w(2+3x)\frac{d\phi}{dx} \right )_{x=l}-\int_{0}^{1}\frac{dw}{dx}(2+3x)\frac{d\phi}{dx}dx-\int_{0}^{1}w(5x)dx=0,\forall w, w(0)=0 $$     (4.4.16)
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$$ \displaystyle w(2+3*1)(-6) -\int_{0}^{1}\frac{dw}{dx}(2+3x)\frac{d\phi}{dx}dx-\int_{0}^{1}w(5x)dx=0 $$     (4.4.17)
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$$ \displaystyle 30w -\int_{0}^{1}\frac{dw}{dx}(2+3x)\frac{d\phi}{dx}dx-\int_{0}^{1}w(5x)dx=0 $$     (4.4.18)
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$$ \displaystyle \int_{0}^{1}\frac{dw}{dx}(2+3x)\frac{d\phi}{dx}dx=30w -\int_{0}^{1}w(5x)dx $$     (4.4.19)
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$$ \displaystyle \int_{0}^{1}\frac{dw}{dx}(2+3x)\frac{d\phi}{dx}dx=30w(x) -\int_{0}^{1}w(x)(5x)dx $$       (4.4.20)
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Part b
From (4.4.23)
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$$ \displaystyle \frac{d}{dx}\left ( (2+3x)\frac{d\phi}{dx} \right )+5x=0 $$     (4.4.21)
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$$ \displaystyle \int d(2+3x)\frac{d\phi}{dx}+\int5xdx=0 $$
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$$ \displaystyle (2+3x)\frac{d\phi}{dx}+ \frac{5x^{2}}{2}+C_{1}=0 $$     (4.4.22)
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$$ \displaystyle \int (2+3x)d\phi= \frac{- \frac{5x^{2}}{2}+C_{1}}{(2+3x)} $$     (4.4.23)
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$$ \displaystyle \phi= \int - \frac{5x^{2}}{2(2+3x)}+\int \frac{C_{1}}{(2+3x)} $$     (4.4.24)
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From 2.4.4
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$$ \displaystyle \int \frac{x^2}{2+3x}dx=\frac{9x^{2}-12x+8log(2+3x)}{54} $$     (4.4.25)  (4.4.24) becomes:
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$$ \displaystyle \phi= -\frac{5}{2}\left (\frac{9x^{2}-12x+8log(2+3x)}{54} \right ) + \frac{C_{1}}{3} {log(2+3x)}+C_{2} $$       (4.4.26) Apply (4.4.15) to (4.4.22)
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$$ \displaystyle (2+3*1)*-6+ \frac{5*1^{2}}{2}+C_{1}=0$$ $$ \displaystyle C_{1}=-27.5 $$     (4.4.27)
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Apply (4.4.27) to (4.4.26)
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$$ \displaystyle \phi= -\frac{5}{2}\left (\frac{9x^{2}-12x+8log(2+3x)}{54} \right ) + \frac{-27.5}{3} {log(2+3x)}+C_{2} $$     (4.4.28)  Apply (4.4.14) to (4.4.28)
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$$ \displaystyle \phi(x=0)=4= -\frac{5}{2}\left (\frac{9*0^{2}-12*0+8log(2+3*0)}{54} \right ) + \frac{-27.5}{3} {log(2+3*0)}+C_{2} $$ $$ \displaystyle C_{2}=6.8709342 $$     (4.4.29)  The final solution is:
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$$ \displaystyle \phi= -\frac{5}{2}\left (\frac{9x^{2}-12x+8log(2+3x)}{54} \right ) + \frac{-27.5}{3} {log(2+3x)}+ 6.8709342 $$       (4.4.26)
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Author
Brandonhua 04:06, 2 March 2011 (UTC)

Problem Description
Possible basis functions are given as:

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

Cosine: $${{F}_{C}}=\left\{ \cos jx,\text{ }j=0,1,2,\ldots \right\}$$

Sine: $${{F}_{S}}=\left\{ 1,\text{ }\sin jx,\text{ }j=1,2,\ldots \right\}$$

Fourier: $${{F}_{F}}=\left\{ \cos jx,\text{ }j=0,1,2,\ldots ;\text{ }\sin kx,\text{ }k=1,2,\ldots \right\}$$

Exponential: $${{F}_{E}}=\left\{ {{e}^{jx}},\text{ }j=0,1,2,\ldots \right\}$$

For $${{\Gamma }_{g}}=\left\{ \beta \right\}$$, and each of the above families of basis functions, i.e., $${{\Gamma }_{I}}=\left\{ {{b}_{j}}(x),\text{ }j=0,1,\ldots \right\},\text{ }I\in \left\{ P,C,S,F,E \right\}$$

Find
Find corresponding family $${{\bar{F}}_{I}}=\left\{ {{{\bar{b}}}_{j}} \right\}$$ satisfying constraint breaking solution (CBS), i.e., $${{b}_{0}}(\beta )\ne 0$$ and $${{b}_{i}}(\beta )=0,\text{ }i=1,2,\ldots ,n$$. Let $$\Omega =\left] \alpha ,\beta \right[=\left] -2,4 \right[$$.

1) $${{\bar{b}}_{0}}(x)=\text{const}$$, plot $${{b}_{j}}$$ and $${{\bar{b}}_{j}}$$ for $$j=1,2,3$$.

2) Show $$\left\{ {{e}^{jx}},\text{ }j=0,1,2 \right\}$$ linear independent on $$\Omega $$.

Part 1
The corresponding family $${{\bar{F}}_{I}}=\left\{ {{{\bar{b}}}_{j}} \right\}$$ satisfying CBS:

Polynomial:

$${{F}_{P}}=\left\{ {{\left( x-4 \right)}^{j}},\text{ }j=0,1,2,\ldots \right\}$$

Cosine:

$${{F}_{C}}=\left\{ 1,\text{ }\cos \left( j\left( x-4 \right)+\frac{\pi }{2} \right),\text{ }j=1,2,\ldots \right\}$$

Sine:

$${{F}_{S}}=\left\{ 1,\text{ }\sin \left( j\left( x-4 \right) \right),\text{ }j=1,2,\ldots \right\}$$

Fourier:

$${{F}_{F}}=\left\{ 1,\text{ }\cos \left( j\left( x-4 \right)+\frac{\pi }{2} \right),\text{ }j=1,2,\ldots ;\text{ }\sin \left( k\left( x-4 \right) \right),\text{ }k=1,2,\ldots \right\}$$

Exponential:

$${{F}_{E}}=\left\{ 1,\text{ }{{e}^{j\left( x-4 \right)}}-1,\text{ }j=1,2,\ldots \right\}$$

The following figures show the polynomial basis functions (j=1,2,3):

The following figures show the cosine basis functions (j=1,2,3):

The following figures show the sine basis functions (j=1,2,3):

The following figures show the fourier basis functions (j=1,2,3):

The following figures show the exponential basis functions (j=1,2,3):

Part 2
The exponential basis function is $${{b}_{j}}=\left\{ {{e}^{jx}},\text{ }j=0,1,2 \right\}$$,

Thus the Gram matrix is:

$$\Gamma =\left[ \begin{matrix} 6 & {{e}^{4}}-{{e}^{-2}} & \frac{1}{2}\left( {{e}^{8}}-{{e}^{-4}} \right) \\ {{e}^{4}}-{{e}^{-2}} & \frac{1}{2}\left( {{e}^{8}}-{{e}^{-4}} \right) & \frac{1}{3}\left( {{e}^{12}}-{{e}^{-6}} \right) \\ \frac{1}{2}\left( {{e}^{8}}-{{e}^{-4}} \right) & \frac{1}{3}\left( {{e}^{12}}-{{e}^{-6}} \right) & \frac{1}{4}\left( {{e}^{16}}-{{e}^{-8}} \right) \\ \end{matrix} \right]$$

And the determinant of Gram matrix is:

$$\det \left( \Gamma \right)=1.1145\times {{10}^{9}}\ne 0$$

Thus, the exponential basis function is linear independent on $$\Omega $$.

Author
Jiang Jin

Problem Description
An elastic bar is subjected along its length to a variable distributed spring p(x) as seen in the figure below. An axial force proportional to the displacement is imposed by the spring on the bar. Adapted from "A First Course in Finite Elements" by Fish and Belytschko p. 74 pb.3.9

a. Construct the strong form.

b. Construct the weak form.

Given
This bar is of length l, cross-sectional area A(x), Young's modulus E(x) with body force b(x) and with boundary conditions:

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$$\displaystyle u(0)=0$$
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$$\displaystyle E(l)A(l)\frac{du(l)}{dx}=t$$
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As seen in the figure above.

Solution
Assumptions:
 * The bar must be in equilibrium
 * It satisfies Hooke's law
 * Compatible Displacement Field
 * The Strain-Displacement Equation must be Satisfied

a. Construct the Strong Form
Following a similar solution method as Team 2 used here with Eq. 1.1-1.10. Using the following diagram from Team 7 homework 1 to start creating the strong form:

Performing a force balance on the rod using the free body diagram above, the net force is equal to the internal forces plus the external, or applied, forces. Note, in this specific problem we will designate our body force as b(x+dx/2) (instead of f(x+dx/2 as seen in the diagram from Team 7) and have the additional body force in the opposite direction caused by the springs (as seen by the figure in the problem description) designated as p(x+dx/2)

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$$ \displaystyle \Sigma \text{ Internal Force } + \text{ Applied Force } = \text{ Net Force}$$ (4.6.1)
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$$ \displaystyle \Sigma \text{ Internal Force }=N(x) + N(x+\Delta x)$$ (4.6.2)
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$$ \displaystyle \text{ Applied Force }=b(x+\frac{dx}{2})dx-p(x+\frac{dx}{2})dx$$ (4.6.3)
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Since we are looking at the static case:

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$$ \displaystyle \text{ Net Force}= m*a =0$$ (4.6.4)
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Substituting equations 4.6.2-4.6.4 into the force balance equation, 4.6.1:

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$$ \displaystyle N(x) + N(x+dx) + b(x+\frac{dx}{2})dx-p(x+\frac{dx}{2})dx =0$$ (4.6.5)
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Plugging in for the direction of the surface normal, the balance of forces at position x can be expressed as:

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$$ \displaystyle N(x)=n(x)\sigma (x)A(x)=-\sigma (x)A(x)$$ (4.6.6)
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Also plugging in for the surface normal, the corresponding force at x + dx:

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$$ \displaystyle N(x+dx)=n(x+dx)\sigma (x+dx)A(x+dx)=\sigma (x+dx)A(x+dx)$$ (4.6.7)
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Using a Taylor Series:

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$$ \displaystyle \sigma (x+dx)=\sigma (x)+\frac{d\sigma (x)}{dx}dx+...$$ (4.6.8)
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$$ \displaystyle A(x+dx)=A(x)+\frac{dA(x)}{dx}dx+...$$ (4.6.9)
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Additionally, in Eq. 4.6.3 the body force $$b(x+\frac{dx}{2})$$ has been averaged for the differential volume we are considering to formulate the governing differential equation. The same has been done for the force caused by the variable distributed spring, given as $$p(x+\frac{dx}{2})$$. Taylor expanding for these as well:

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$$\displaystyle b(x+\frac{dx}{2})=b(x)+\frac{db(x)}{dx}\frac{dx}{2}....$$. (4.6.10)
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$$\displaystyle p(x+\frac{dx}{2})=p(x)+\frac{dp(x)}{dx}\frac{dx}{2}....$$. (4.6.11)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

Plugging in Eq.'s 4.6.8 and 4.6.9 into Eq. 4.6.7, and then plugging this and Eq.'s 4.6.6, 4.6.10 and 4.6.11 into Eq. 4.6.5 gives:
 * {| style="width:100%" border="0"

$$\displaystyle \sigma (x)A(x) + \sigma (x)\frac{dA(x)}{dx}dx+A(x)\frac{d\sigma (x)}{dx}-\sigma (x)A(x)+b(x)dx-p(x)dx$$. (4.6.12)
 * style="width:10%; |
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 *

 * }

Using the product rule, canceling out terms, and dividing through by dx gives:
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$$\displaystyle \frac{d}{dx}\sigma (x)A(x)+b(x)-p(x)=0$$. (4.6.13)
 * style="width:10%; |
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 *

 * }

Noting that the strain-displacement equation is
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$$ \displaystyle
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\varepsilon \left( x \right) = \frac

$$     (4.6.14) And by Hooke ‘ s Law
 *

 * }
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$$ \displaystyle
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\sigma \left( x \right) = E\left( x \right)\varepsilon \left( x \right) = E\left( x \right)\frac

$$     (4.6.15)
 *

 * }

Gives the following strong form:
 * {| style="width:92%; padding:10px; border:2px solid #8888aa"

$$\displaystyle \frac{d}{dx}E(x)A(x)\frac{du}{dx}+b(x)-p(x)=0$$. (4.6.16)
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 *

 * }

b. Construct the Weak Form
Following the same solutions as Team 7's previous work here.

In order to reduce the requirements or constraints on what trial solutions could be substituted in for u, we will generate the weak form which requires the basis solutions only be once differentiable. This occurs from the natural boundary condition being absorbed in the weak form, as will be seen shortly. Also, this creates a symmetric stiffness matrix.

As with Team 7's solution, the strong form and natural boundary condition are used in an inner product with an arbitrary weighting function giving the following equations:

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$$\displaystyle \int_{0}^{l}w(x)\big [\frac{d}{dx}E(x)A(x)\frac{du}{dx}+b(x)-p(x)\big ]dx=0$$. (4.6.17)
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 *

 * }

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$$\displaystyle w(l)E(l)A(l)\frac{du(l)}{dx}-w(l)t=0$$. (4.6.18)
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 *

 * }

Distributing the weighting function w(x) and the integral of Eq. 4.6.17 and looking at the first term:

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$$\displaystyle \int_{0}^{l}w\frac{d}{dx}E(x)A(x)\frac{du}{dx}dx=w(x)E(x)A(x)|_{0}^{l}-int_{0}^{l}\frac{dw}{dx}E(x)A(x)\frac{du}{dx}dx$$. (4.6.19)
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 *

 * }

In Eq. 4.6.19 integration by parts (derived and example of use here) was used. Now to remove the unknown flux at the essential boundary location (x=0), we will make the arbitrary weighting function be zero at x equal to zero, or

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$$\displaystyle w(0)=0$$. (4.6.20) Using Eq. 4.6.7 and 4.6.9, Eq. 4.6.8 becomes:
 * style="width:10%; |
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 *

 * }

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$$\displaystyle w(x)E(x)A(x)|_{0}^{l}-\int_{0}^{l}\frac{dw}{dx}E(x)A(x)\frac{du}{dx}dx=w(l)t-\int_{0}^{l}\frac{dw}{dx}E(x)A(x)\frac{du}{dx}dx$$. (4.6.21)
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 *

 * }

Combining Equations 4.6.18 and 4.6.21 gives the following weak form:

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$$\displaystyle w(l)t-\int_{0}^{l}\frac{dw}{dx}E(x)A(x)\frac{du}{dx}dx+\int_{0}^{l}w(x)p(x)dx+\int_{0}^{l}w(x)b(x)dx$$. (4.6.22)
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 *

 * }

Note in equation 4.6.22 the weighting function must be zero at the essential boundary condition (w(0)=0). Also, the weighting function and the displacement are both first order derivatives and now a symmetric stiffness matrix can be formed.

Authors
Braden Snook, Zongyi Yang

Problem Description
A calculix manual for novice/dummies

Author
Eml5526.s11.team7.Srilalithkumar

Problem Description
Consider a trial solution of the form $$u(x) = {\alpha _0} + {\alpha _1}(x - 3) + {\alpha _2}{(x - 3)^2} $$and a weight function of the same form. Obtain a solution to the weak form in Problem 3.1, "A First Course in Finite Elements", by Fish & Belytschko P72. Find $$\tilde M$$, assuming $$A = 1,E = 2,\bar m = 3$$, Do for $$n = 3$$. For dynamics, with $${u^h}(\beta ,t) = g(t) = \sin 2t$$, find $$F(t)$$ The weak form of Problem 3.1 is
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$$ \displaystyle
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\int\limits_1^3 {\fracAE\frac} dx = - 0.1{(wA)_{x = 1}} + \int\limits_1^3 {2xwdx\forall w(3) = 0}

$$     (4.8.1)
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$$ \displaystyle
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u(3) = 0.001

$$
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 * }

Solution
Part 1: The basis functions $${b_i} = \{ 1,(x - 3),{(x - 3)^2},{(x - 3)^3}\} i = 0,1,2,3$$
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$$ \displaystyle
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\tilde M = \left[ {\begin{array}{ccccccccccccccc} & \\  & \end{array}} \right]

$$     (4.8.2)
 *

 * }
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$$ \displaystyle
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{M_{EE}} = {\tilde M_{00}} = \tilde m({b_0},{b_0}) = \int\limits_1^3 {{b_0}\bar m{b_0}dx} = \int\limits_1^3 {3dx}  = 6

$$     (4.8.3)
 *

 * }
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$$ \displaystyle
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{M_{EF}} = \left[ {{M_{0j}};j = 1,2,3} \right] = \tilde m({b_0},{b_j})

$$     (4.8.4)
 *

 * }
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$$ \displaystyle
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\tilde m({b_0},{b_1}) = \int\limits_1^3 {{b_0}\bar m{b_1}dx} = \int\limits_1^3 {3(x - 3)dx}  =  - 6

$$     (4.8.5)
 *

 * }
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$$ \displaystyle
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\tilde m({b_0},{b_2}) = \int\limits_1^3 {{b_0}\bar m{b_2}dx} = \int\limits_1^3 {3{{(x - 3)}^2}dx}  = 8

$$     (4.8.6)
 *

 * }
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$$ \displaystyle
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\tilde m({b_0},{b_3}) = \int\limits_1^3 {{b_0}\bar m{b_3}dx} = \int\limits_1^3 {3{{(x - 3)}^3}dx}  =  - 12

$$     (4.8.7)
 *

 * }
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$$ \displaystyle
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{M_{EF}} = \left[ {\begin{array}{ccccccccccccccc} { - 6}&8&{ - 12} \end{array}} \right]

$$     (4.8.8)
 *

 * }
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$$ \displaystyle
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{M_{FE}} = {M_{EF}}^T = \left[ {\begin{array}{ccccccccccccccc} { - 6} \\  8 \\   { - 12} \end{array}} \right]

$$     (4.8.9)
 *

 * }
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$$ \displaystyle
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{M_{FF}} = \left[ {{M_{ij}};i,j = 1,2,3} \right] = \tilde m({b_i},{b_j})

$$     (4.8.10)
 *

 * }
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$$ \displaystyle
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\tilde m({b_1},{b_1}) = \int\limits_1^3 {{b_1}\bar m{b_1}dx} = 3\int\limits_1^3 {(x - 3)(x - 3)dx}  = 8

$$     (4.8.11)
 *

 * }
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$$ \displaystyle
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\tilde m({b_1},{b_2}) = \int\limits_1^3 {{b_1}\bar m{b_2}dx} = 3\int\limits_1^3 {(x - 3){{(x - 3)}^2}dx}  =  - 12

$$     (4.8.12)
 *

 * }
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$$ \displaystyle
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\tilde m({b_1},{b_3}) = \int\limits_1^3 {{b_1}\bar m{b_3}dx} = 3\int\limits_1^3 {(x - 3){{(x - 3)}^3}dx}  = \frac{5}

$$     (4.8.13)
 *

 * }
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$$ \displaystyle
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\tilde m({b_2},{b_1}) = \int\limits_1^3 {{b_2}\bar m{b_1}dx} = 3\int\limits_1^3 {{{(x - 3)}^2}(x - 3)dx}  =  - 12

$$     (4.8.14)
 *

 * }
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$$ \displaystyle
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\tilde m({b_2},{b_2}) = \int\limits_1^3 {{b_2}\bar m{b_2}dx} = 3\int\limits_1^3 {{{(x - 3)}^2}{{(x - 3)}^2}dx}  = \frac{5}

$$     (4.8.15)
 *

 * }
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$$ \displaystyle
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\tilde m({b_2},{b_3}) = \int\limits_1^3 {{b_2}\bar m{b_3}dx} = 3\int\limits_1^3 {{{(x - 3)}^2}{{(x - 3)}^3}dx}  =  - 32

$$     (4.8.16)
 *

 * }
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$$ \displaystyle
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\tilde m({b_3},{b_1}) = \int\limits_1^3 {{b_3}\bar m{b_1}dx} = 3\int\limits_1^3 {{{(x - 3)}^3}(x - 3)dx}  = \frac{5}

$$     (4.8.17)
 *

 * }
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$$ \displaystyle
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\tilde m({b_3},{b_2}) = \int\limits_1^3 {{b_3}\bar m{b_2}dx} = 3\int\limits_1^3 {{{(x - 3)}^3}{{(x - 3)}^2}dx}  =  - 32

$$     (4.8.18)
 *

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

\tilde m({b_3},{b_3}) = \int\limits_1^3 {{b_3}\bar m{b_3}dx} = 3\int\limits_1^3 {{{(x - 3)}^3}{{(x - 3)}^3}dx}  = \frac{7}

$$     (4.8.19)
 *

 * }
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$$ \displaystyle
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\therefore {M_{FF}} = \left[ {\begin{array}{ccccccccccccccc} 8&{ - 12}&{\frac{5}} \\ { - 12}&{\frac{5}}&{ - 32} \\ {\frac{5}}&{ - 32}&{\frac{7}} \end{array}} \right]

$$     (4.8.20)
 *

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

\therefore {M_{FF}} = \left[ {\begin{array}{ccccccccccccccc} 6&{ - 6}&8&{ - 12} \\  { - 6}&8&{ - 12}&{\frac{5}} \\ 8&{ - 12}&{\frac{5}}&{ - 32} \\ { - 12}&{\frac{5}}&{ - 32}&{\frac{7}} \end{array}} \right]

$$     (4.8.21) Part 2:
 *

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$$ \displaystyle
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F = {F_F} - \left( {{M_{FE}}{g^{(2)}} + {K_{FE}}g} \right)

$$     (4.8.22)
 *

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

$$     (4.8.23)
 *

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

$$     (4.8.24)
 *

 * }
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$$ \displaystyle
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\because f = 0

$$     (4.8.25)
 *

 * }
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$$ \displaystyle
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\therefore \tilde f({b_i}) = {b_i}(1)h

$$     (4.8.26)
 *

 * }
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$$ \displaystyle
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h = {a_2}(\alpha )\frac = AE\frac = 0.1

$$     (4.8.27)
 *

 * }
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$$ \displaystyle
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\tilde f({b_1}) = {b_1}(1)h = - 2 \times 0.1 =  - 0.2

$$     (4.8.28)
 *

 * }
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$$ \displaystyle
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\tilde f({b_2}) = {b_2}(1)h = {\left( { - 2} \right)^2} \times 0.1 = 0.4

$$     (4.8.29)
 *

 * }
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$$ \displaystyle
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\tilde f({b_3}) = {b_3}(1)h = {\left( { - 2} \right)^3} \times 0.1 = - 0.8

$$     (4.8.30)
 *

 * }
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$$ \displaystyle
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\therefore {F_F} = \left[ {\begin{array}{ccccccccccccccc} { - 0.2} \\  {0.4} \\   { - 0.8} \end{array}} \right]

$$     (4.8.31)
 *

 * }
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$$ \displaystyle
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{g^{(2)}} = - 4\sin (2t)

$$     (4.8.32)
 *

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

$$     (4.8.33)
 *

 * }
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$$ \displaystyle
 * style="width:95%" |
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\because \tilde k({b_i},{b_j}) = \int\limits_1^3 {b{'_i}{a_2}b{'_j}dx} = \int\limits_1^3 {b{'_i}AEb{'_j}dx}

$$     (4.8.34) So
 *

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

\tilde k({b_0},{b_1}) = \int\limits_1^3 {b{'_0}{a_2}b{'_1}dx} = \int\limits_1^3 {b{'_0}AEb{'_1}dx}  = 0 = \tilde k({b_0},{b_2}) = \tilde k({b_0},{b_3}) = 0

$$     (4.8.35)
 *

 * }
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$$ \displaystyle
 * style="width:95%" |
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\therefore {K_{EF}} = \left[ {\begin{array}{ccccccccccccccc} 0 \\  0 \\   0 \end{array}} \right]

$$     (4.8.36) And：
 *

 * }
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$$ \displaystyle
 * style="width:95%" |
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\because {M_{FE}} = {M_{EF}}^T = \left[ {\begin{array}{ccccccccccccccc} { - 6} \\  8 \\   { - 12} \end{array}} \right]

$$     (4.8.37)
 *

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

F = {F_F} - \left( {{M_{FE}}{g^{(2)}} + {K_{FE}}g} \right) = \left[ {\begin{array}{ccccccccccccccc} { - 0.2 - 24\sin (2t)} \\ {0.4 + 32\sin (2t)} \\ { - 0.8 - 48\sin (2t)} \end{array}} \right]

$$     (4.8.38)
 *

 * }

Authors
Zongyi Yang

Jiang Jin

title: User:Eml5526.s11.team7/HW4


text: File:Ploynomial.jpg

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text:
User:Egm6341.s11.team3/sub4

= 4.1 Verify the accuracy of Simpson's rule =

Given: a generic 3rd order polynomial

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$$f(x) = p_3(x) = c_0 + c_1x + c_2x^2 + c_3x^3\ $$ $$(Eq.4.1.1) \ $$
 * style="width:95%" |
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 *
 * }

with

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$$c_0=1,\ c_1=3,\ c_2=-9,\ c_3=12$$
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 *
 * }

Find: approximate and exact solutions
$$I\ $$ the exact integration on $$[-2,\ 1] $$

$$I_2\ $$ the numerical integration using simple Simpson's rule on $$[-2,\ 1] $$

Solution

 * Exact solution

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

$$I=\int_{-2}^{1} f(x) = \int_{-2}^{1} (c_0 + c_1x + c_2x^2 + c_3x^3) dx= [c_0 x]_{-2}^{1}+\left[\frac{c_1}{2}x^2\right]_{-2}^{1}+\left[\frac{c_2}{3}x^3\right]_{-2}^{1}+\left[\frac{c_3}{4}x^4\right]_{-2}^{1}=-73.5\ $$ $$ (Eq.4.1.2) \ $$
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 *
 * }

 * Numerical solution:

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$$f_0=f(-2)=c_0-2c_1+4c_2-8c_3=-137\ $$ $$ (Eq.4.1.3) \ $$
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 *
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$$f_1=f(-0.5)=c_0-0.5c_1+0.25c_2-0.125c_3=-4.25\ $$ $$ (Eq.4.1.4) \ $$
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 *
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$$f_0=f(1)=c_0+c_1+c_2+c_3=7\ $$ $$ (Eq.4.1.5) \ $$
 * style="width:95%" |
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 *
 * }

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$$h=\dfrac{1-(-2)}{2}=1.5\ $$ $$ (Eq.4.1.6) \ $$ Using simple Simpson's rule
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 *
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 * {| style="width:100%" border="0"

$$I_2=\dfrac{h}{3}(f_0+4f_1+f_2)=-73.5=I\ $$ $$ (Eq.4.1.7) \ $$
 * style="width:95%" |
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 *
 * }

Egm6341.s11.team3.ren 19:23, 1 March 2011 (UTC)

title: User:Egm6341.s11.team3/sub4


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title: File:Sine.jpg


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title: File:Fourier j=3.jpg


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text: File:Fe1.s11.tm7.HW4.fig4.4.1 22.jpg

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