text:
User:EGM6341.s11.TEAM1.HW6

Problem 1: The Theorem of Higher Order Trapezoidal Rule Error (HOTRE) continued
 Solved without assistance

Perform steps 4ab to find $$ (p_6,p_7) \ $$ and $$ E \ $$
Given: Steps 3ab:

Step 4b - Find $$ p_7(t) \ $$ by integration by parts
Next, we want $$ (c_7,c_8) \ $$ to cancel terms with $$ q^{(5)} \ $$ even order derivative of q. Set $$ p_7(t) = 0 \ $$, such that $$ p_7(0) = 0 \ $$ and $$ p_7( \pm \ 1) = 0 \ $$ Using given values for $$ c_1, c_3, c_5 \ $$ and solving for $$ p_7(0) = 0 \ $$ gives: Which results in $$ c_8=0 \ $$ Using given values for $$ c_1, c_3, c_5 \ $$ and solution of $$ c_8 \ $$, now solve for $$ p_7(1) = 0 \ $$ gives: Which results in $$ c_7 = \frac{31}{15120} \ $$, verify for $$ p_7(-1) = 0 \ $$: Again results in $$ c_7 = \frac{31}{15120} \ $$

$$ \therefore p_6(t) = -\frac{t^6}{6!}+ \frac{1}{6}\frac{t^4}{4!}- \frac{7}{360}\frac{t^2}{2!}+\frac{31}{15120} \ $$ $$ \therefore p_7(t) = -\frac{t^7}{7!}+ \frac{1}{6}\frac{t^5}{5!}- \frac{7}{360}\frac{t^3}{3!}+\frac{31}{15120} \ $$

Find $$ t_k(x) \ $$ from Eq(3) p30-2
 Referenced from User:Egm6341.s10.Team4/HW5: in order to understand re-arrangement of terms

From p30-2 Eq(3): for $$ t \in \left [ -1,+1 \right ] \ $$

Now substitute $$ x=x_k \ $$

Knowing that $$ x_{k+1}-x_k=h \ $$

Now substitute $$ x=x_{k+1} \ $$

Knowing that $$ x_{k+1}-x_k=h \ $$

$$ \therefore \ $$ When $$ x=x_k \ $$, $$ t_k(x)=-1 \ $$ and $$ \therefore \ $$ When $$ x=x_{k+1} \ $$, $$ t_k(x)=1 \ $$

Problem 2: Derive Eq(1) p31-1 and Eq(2) p31-1
 Solved without assistance

Derive Eq(1) p31-1 $$ E_n^T \ $$
Derive the Composite Trapezoidal Rule Error Equation

The error of Comp. Trap. Rule

Transfer $$ \int_{x_{k}}^{x_{k+1}}f(x)dx \ $$ to $$ \int_{-1}^{+1}g_{k}(t)dt \ $$

Hence,

Applying successive integration by parts, the following equation will be obtained:

Since $$ p_{2r} \ $$ is an even function, Take higher derivation of $$ g_k(t) \ $$, Substitute (2.5) and (2.6) into (2.4), $$ E \ $$ can be rewritten as follows: Let $$ x_k=a, x_{k+1}=b \ $$and rearrange the equation, Setting $$ \bar{d_{2r}}=\frac{p_{2r}(+1)}{2^{2r}} \ $$, $$ E \ $$ can be obtained as the following equation.

Derive Eq(2) p31-1 $$ \bar d \ _{2r} \ $$
Shown in the above, $$ \bar{d}_{2r} \ $$ can be obtained from (2.8).

Problem 3: Given Bernoulli's Numbers $$ x \coth x= \sum_{r=0}^{ \infty } \frac{B_{2r}}{(2r)!} x^{2r} \ $$
 Solved without assistance

Verify $$ \bar d \ _2,\bar d \ _4, \bar d \ _6 \ $$
Above result is exactly same with $$\displaystyle \bar{d_{2r}} = \frac{P_{2r}(1)}{2^{2r}}$$

Result table

Find $$ \bar d \ _8,\bar d \ _{10} $$ using $$ x \coth x= \sum_{r=0}^{ \infty } \frac{B_{2r}}{(2r)!} x^{2r} \ $$
Since $$\displaystyle \bar{d_{2r}} = \frac{P_{2r}(1)}{2^{2r}}$$,

Result table

Find $$ \bar d \ _8,\bar d \ _{10} $$ using Bernoulli nos. and compare to solution found in 3.2
Then, based on the relation between $$\displaystyle \bar{d_{2r}} $$ and Bernoulli numbers, we can obtain $$\displaystyle \bar{d_{2}}, \,\bar{d_{4}}, \,\bar{d_{6}}$$ as follows.

Since $$\displaystyle B_{2}=\frac{1}{6}, \, B_{4}=-\frac{1}{30}, \, B_{6}=\frac{1}{42}, B_{8}=-\frac{1}{30}, \, B_{10}=\frac{5}{66}$$,

Result table

Problem 4: Redo proof of Higher Order Trapezoidal Rule Error(HOTRE) (ref: HW5, p8) by trying to cancel terms with odd order of derivative of g
 Solved without assistance

Problem objective
Redo proof of Higher Order Trapezoidal Rule Error(HOTRE) (ref: HW5, p8) by trying to cancel terms with odd order of derivative of g

The composite trap error is given as:

Integrate E using integration by part we have:

Solution
At $$\displaystyle t = \pm 1$$, we get $$ c_3, c_4, c_5 \ $$

Expanding the error eqn with integration by parts,

We can cancel odd order derivatives of g and find the general expression of E as below
 * {| style="width:100%" border="0"

Note:Characteristic of odd func. $$\begin{align} \left[ P_{2l+1}(t)g^{(2l)}(t) \right]_{-1}^{+1} &= \underbrace{P_{2l+1}(+1)}_{\text{odd func.}}g^{(2l)}(+1)-\underbrace{P_{2l+1}(-1)}_{\text{odd func.}}g^{(2l)}(-1)\\ &=- P_{2l+1}(-1)\left[(g^{(2l)}(+1)+g^{(2l)}(-1))\right]\\ \end{align}$$ (4.9)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

We get,

Using the given relation,

We get,

Plug the above eqns into the general form,

Problem 5: Given $$ p_1(t)=-t \ $$, use $$ \frac{c_1}{(2i+1)!}+\frac{c_3}{(2i-1)!}+\frac{c_5}{(2i-3)!}+...+\frac{c_{2i-1}}{3!}+c_{2i+1}=0 \ $$ to find 5.1, 5.2, 5.3 and 5.4
 Solved without assistance

Using problem statement Find $$ (p_2, p_3) \ $$
Using the Recurrence formula on p.32-2(1) to obtain the pairs of polynomials $$ (p_2, p_3),(p_4, p_5),(p_6,p_7),(p_8,p_9) \ $$. The Recurrence formula are as follows,

We know $$ p_1(t)=-t \ $$ for $$ p_1 \ $$, we can get $$ i=0 \ $$

$$ C_1=-1 \ $$ for $$ (p_2,p_3) \ $$, we can get $$ i=1 \ $$

Using problem statement Find $$ (p_4, p_5) \ $$
Similarly, for $$ (p_4,p_5) \ $$, we can get $$ i=2 \ $$,

Using problem statement Find $$ (p_6, p_7) \ $$
for $$ (p_6, p_7) \ $$, we can get $$ i=3 \ $$

Using problem statement Find $$ (p_8, p_9) \ $$
for $$ (p_8, p_9) \ $$, we can get $$ i=4 \ $$,

Problem 6: Reproduce Kessler's Results using $$ (p_{2i}, p_{2i+1}) \ $$, where $$ i=1,2,3 \ $$. Start with the best of S10 to understand Kessler's code and reproduce its result
 Referenced from S10,Egm6341 Team1 HW5 Problem#10 to understand Kessler's Matlab code:|S10,Team1(HW5), but revised it to obtain correct P's and C's with s11 values

Reproduce Kessler's results
From eqn.6.1, 6.2 and 6.3, we can have following table;

By using the Kessler's Matlab code, n=3, following two tables are obtained;

In problem.5, we obtained $$C_3, C_5 \ $$ and $$ C_7 \ $$ using eqn.6.1, 6.2, and 6.3; And also, we can compute $$ p_2(1) \ $$ through $$ p_7(1) \ $$ using equations in the first above table, and from lecture note [[media:Nm1.s11.mtg32.djvu|(3) 32-1]], we know that $$ p_{2i+1}(1)=0 \ $$; Let's compare these c's and p(1)'s with that from Kessler's code in the second and third above table. $$\Rightarrow \;$$ c's are totally same with each result, but p(1)'s are different. So, it is needed to revise Kessler's code in order to generate correct p(1)'s. To be more specific, function [nsum, dsum]=fracsum(n,d) works correctly for updating cn and cd variables with proper values, but has flaws to generate correct coefficients, p(1)'s, which are considered as a sum of input vectors in this function. Followings are (1) new function, [nsum, dsum]=re_fracsum(n,d) for newpn and newpd variables and (2) partially changed function, [c,p]=traperror(n).;

Finally, correct coefficients, p(1)'s are obtained from changed Kessler's code as following; , $$\begin{align}\;p_1(1)=p_3(1)=p_5(1)=p_7(1)=0\end{align}$$

Show that $$ z_{i+\frac{1}{2}} = z \left ( s=\frac{1}{2} \right ) = \frac{1}{2}(z_i+z_{i+1})+\frac{h}{8}(f_i-f_{i+1}) $$
From lecture note [[media:Nm1.s11.mtg37.djvu|p.37-3]], we know coefficients, $$\displaystyle\begin{align}c_i's\end{align}$$. From eqn.6.4, each $$\displaystyle C_i $$ term can be replaced with $$\displaystyle Z $$ term. From lecture note, [[media:Nm1.s11.mtg37.djvu|(6) p.37-1]] and [[media:Nm1.s11.mtg38.djvu|(4) p.38-1]], we know $$\displaystyle\dot{Z_i}=Z_i'\frac{1}{h}=f_i$$. Thus,

Show that $$ Z'_{i+1/2}=Z'(s=1/2)=-\frac{3}{2}(Z_i-Z_{i+1})-\frac{1}{4}(Z'_i+Z'_{i+1}) $$
From lecture note [[media:Nm1.s11.mtg37.djvu|(1) p.37-2]], we know that Using eqn.6.4, matrix, each $$\displaystyle C $$ can be represented in $$\displaystyle Z $$ term. Thus,

Problem 7: Methods of computing cicumferences
Use Wolfram Alpha for reference Provide elapsed time of you code in your solutions Use MATLAB command tic and tac

 Solved without assistance   Referenced [link here], but solved with s11 values

bifolium
solution here

ellipse, where $$ a=10 \ $$ and $$ b=3 \ $$ using $$ r( \theta\ )=\frac{a(1-e^2)}{1-e \cos \theta\ } \ $$ and $$ C = \int_{ \theta\ =0}^{2 \pi\ } dl \ $$
solution here

ellipse where $$ a=10 \ $$ and $$ b=3 \ $$ using $$ E(e) = \int_{0}^{\frac{ \pi\ }{2}} \left [ 1-e^2 \sin \theta\ \right ] ^{\frac{1}{2}}, d \theta\ \ \ $$ and $$ C = 4AE(e) \ $$
solution here

bifolium
solution here

ellipse, where $$ a=10 \ $$ and $$ b=3 \ $$ using $$ r( \theta\ )=\frac{a(1-e^2)}{1-e \cos \theta\ } \ $$ and $$ C = \int_{ \theta\ =0}^{2 \pi\ } dl \ $$
solution here

ellipse where $$ a=10 \ $$ and $$ b=3 \ $$ using $$ E(e) = \int_{0}^{\frac{ \pi\ }{2}} \left [ 1-e^2 \sin \theta\ \right ] ^{\frac{1}{2}}, d \theta\ \ \ $$ and $$ C = 4AE(e) \ $$
solution here

bifolium
solution here

ellipse, where $$ a=10 \ $$ and $$ b=3 \ $$ using $$ r( \theta\ )=\frac{a(1-e^2)}{1-e \cos \theta\ } \ $$ and $$ C = \int_{ \theta\ =0}^{2 \pi\ } dl \ $$
solution here

ellipse where $$ a=10 \ $$ and $$ b=3 \ $$ using $$ E(e) = \int_{0}^{\frac{ \pi\ }{2}} \left [ 1-e^2 \sin \theta\ \right ] ^{\frac{1}{2}}, d \theta\ \ \ $$ and $$ C = 4AE(e) \ $$
solution here

bifolium
solution here

ellipse, where $$ a=10 \ $$ and $$ b=3 \ $$ using $$ r( \theta\ )=\frac{a(1-e^2)}{1-e \cos \theta\ } \ $$ and $$ C = \int_{ \theta\ =0}^{2 \pi\ } dl \ $$
solution here

ellipse where $$ a=10 \ $$ and $$ b=3 \ $$ using $$ E(e) = \int_{0}^{\frac{ \pi\ }{2}} \left [ 1-e^2 \sin \theta\ \right ] ^{\frac{1}{2}}, d \theta\ \ \ $$ and $$ C = 4AE(e) \ $$
solution here

===Problem 8: Derive $$ \widehat{PQ} = \int_{ \theta\ _P}^{ \theta\ _Q} \left [ r^2+ \left ( \frac{dr}{d \theta\ } \right ) ^2 \right ] ^{\frac{1}{2}}\, d \theta\ \ $$ using triangle OAB and the law of cosines===

 Referenced, to understand Taylor series implementation

From http://www.wolframalpha.com/, the law of cosines is defined as:

Where $$ a= r \ $$ length of first side, Where $$ b= r+dr \ $$ length of second side, Where $$ c= dl \ $$ length of third side, Where $$ \alpha\ = d \theta\ \ $$ angle opposite third side, The law of cosines can be applied to the triangle OAB with the sides defined as $$ r \ $$, $$ r+dr \ $$, and inner angle $$ d\theta\ \ $$ as shown above to find the segment $$ AB=dl \ $$.

Assume angle $$ d \theta\ \ $$ is small and substitute $$ \cos d \theta = 1-\frac{d \theta\ ^2}{2} \ $$

The Taylor series expansion of $$ \cos x \ $$ is,

Taking the first two terms of the expansion and applying them to the Eq(8.2) yields:

Since the differential portion $$ drd \theta\ ^2 \ $$ in Eq(8.4) goes to zero more rapidly than $$ dr^2 \ $$ and $$ d \theta\ ^2 \ $$, it is set to zero. The remainder is:

Integrating from $$ \theta\ _P \ $$ to $$ \theta\ _Q \ $$ yields:

Problem 9: Prove matrix $$ \underline{A} \ $$ multiplied with matrix $$ \left \{ c_j \right \} \ $$ gives the matrix $$ \begin{Bmatrix} z_i \\ z_i' \\ z_{i+1} \\ z_{i+1}' \end{Bmatrix} \ $$
 Solved without assistance

Verify Matrix $$ A^{-1} $$
In lecture note [[media:Nm1.s11.mtg37.djvu|p.37-2]], A matrix is defined as following; By using matlab command 'inv', we can easily obtain inverse matrix

Matlabcode:

Identify basis functions $$ \overline{N} \ _i(s) $$ for $$ i=1,2,3,4 $$
In lecture note [[media:Nm1.s11.mtg38.djvu|(2) p.38-1]] and [[media:Nm1.s11.mtg37.djvu|(1) p.37-1]], $$\displaystyle Z(s)$$ is defined as following ; for i=1, 2, 3, and 4 and from lecture note [[media:Nm1.s11.mtg37.djvu|(2) p.37-3]], we know From eqn.6.4 matrix, we have following equations; Now, we can expand eqn.9.2, and find $$ \overline{N}_i,$$, for i=1,2,3,4, in terms of s Thus,

Plot $$\displaystyle N_i(s) $$ for i=1,2,3,4 Matlab Code:

=Contributing Members & Referenced Lecture=

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

title: User:EGM6341.s11.TEAM1.HW6


text:
User:EGM6341.s11.TEAM1.HW7

Problem 0: Rework Cauchy Distribution part of 5.2 and 5.4
Solved without assistance

Given: Linear State Space Model
Given the following:

Find: Model plot with various random noise
A.) Run model and plot $$\underline{x} _0 \ $$ B.) Find the equilibrium point as $$ \lim_{k \to \infty} \underline{x} _{k+1} = \lim_{k \to \infty} \underline{F} ^{k+1} \underline{x} _0 =: \underline{\hat{a} } $$ C.) Use MATLAB randn to generate $$ \begin{Bmatrix} w_j \end{Bmatrix} $$, for $$ j=0,1,2,... \ $$ and plot $$ \begin{Bmatrix} \underline{x} _j \end{Bmatrix} $$, for $$ j=0,1,2,... \ $$. Use a big blue dot for $$ j=0 \ $$, use small dots for the other points. All for $$ \alpha\ = 0.5 $$, $$ \alpha\ = 1 $$  and  $$ \alpha\ = 2 $$ D.) Run model with Cauchy random noise and plot $$\underline{x} _0 \ $$

Run linear state space model
The following MATLAB (v2009b) code was developed to evaluate the state space model until an equilibrium point was reached. The results are shown in Figure 5.2.1.

 * Nm1.s11.team1.HW4.fig5.2.1.png

Find equilibrium point
In order to determine the equilibrium point, the Euclidean distance between the next $$x_{n+1} \ $$ and $$ x_{n} \ $$ was calculated for each step in the model. In this case, the Euclidean distance is defined as:

When this distance became less than $$ 10^{-6} \ $$, the equilibrium point was taken as "found".

The MATLAB code presented above displays the equilibrium point, which is also plotted in Figure 5.2.1 above.

Cauchy Random Noise
The following MATLAB (v2009b) code was developed to evaluate the state space model for 1000 iterations, since an equilibrium point was not reached. The results are shown in Figure 5.2.2.

 * Nm1.s11.team1.HW4.fig5.2.3NEW.png

Problem 4: Spring and Dampener system
Solved without assistance

Find: Equation of the motion and control scheme
A.) Equation of motion B.) State space model of system C.) $$c_{cr}$$ D.) Plot $$x_k \ $$ for $$c=\frac{1}{2}c_{cr},c_{cr},\frac{3}{2}c_{cr} $$, and with Gaussian and Cauchy noise.

Equation of motion
The system(Spring-mass-damper) equation can be defined as,

Each of the elements are

Plugging each of the elements in to the eq4.1 becomes EOM(Equation of the motion) as

Define state space model
State space

Applying the given expression on p29-7, Eq(4):

Rewriting to discrete system equation, as

where I is the identity matrix, and A, B, and h are defined as above

Critical damping
Frequency and damping ratio are given as

Thus,

Part3. With Cauchy noise

 * [[File:Nm1 s11 team1 HW7 p0.png|1000px|center|thumb|Figure 7.0.1: [x1,x2]' trajectory and x1,x2 vs time response]]

Problem 1: Use Hermite Simpson algorithm to integrate $$ \frac{dx}{dt}=rx \left ( 1-\frac{x}{x_{max}} \right ) \ $$
Referenced last years work

Given
$$ x_{max}=15 \ $$ $$ r=1.4 \ $$ 2 cases: $$ x_0=3 < \frac{1}{2}x_{max} \ $$ and $$ x_0=9 > \frac{1}{2}x_{max} \ $$ for $$ t \in [0,10] \ $$

case 1: $$ x_0=3 \ $$
 * Hw7 p2 team 1.png

case 2: $$ x_0=9 \ $$
 * Hw7 p2 team 1 1.png

Problem 2: Solve parameters for Equation of Motion
Referenced the work of Team 3, Spring 2010.

Improvements: -Added display of absolute error to user can monitor progress of convergence -Code restructured, variables renamed, and additional comments and to make code easier to follow and inputs easier to change -Improved parts of code to reduce time for absolute error to converge to $$10^{-6}$$ -->Old code time: Longer than 1 hour -->This code time: 107 sec

Given
Equations of motion for an aircraft:

where

and

Physical parameters were taken from the Subchan and Zbikowski paper:

Initial conditions were taken from the Subchan and Zbikowski paper:

Values for the controls are given in [[media:Egm6341.s10.mtg41.djvu|S10 Mtg 41]]: $$ \underline{u} (t)= \left \lfloor T(t), \alpha\ (t) \right \rfloor \ $$

Find: z and J
Solve to find the matrix $$z \ $$ for the bunt maneuver using S10 and S11 values, as well as values for $$ J \ $$ such that:

$$ J= \int_{0}^{t_f} y(t)\, dt \ $$ such that $$ y(t_f)=0 \ $$

Solve: Using S10 values
The following MATLAB code was developed to solve for $$\underline{z} \ $$ using Newton-Raphson and Hermite-Simpson:

The solution was confirmed using the MATLAB command ode45:

Solve: Using S11 values
The different values of $$\alpha \ $$ used for the S11 solution are shown below:

The following MATLAB code executes the model using these new values of $$\alpha \ $$:

Solve: Compare J values
The following MATLAB code was used to use the MATLAB trapz command to integrate the output of Parts 1 and 2 and determine $$ J_1 \ $$ and $$ J_2 \ $$.

$$J_1 = 16860 \ $$ $$J_2 = 10824 \ $$

Meaning that since $$ J_2 < J_1 \ $$, the S11 values provide a better solution.

Problem 3: Solving the Logistic Equation Using Inconsistent Trap-Simpson's rule & Newton-Raphson
Solved without assistance

Given
Refer Lecture slide 41-1 for problem statement

The logistic equation for population dynamics is given by

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

$$\displaystyle \dot{x}(t) = f(x) = rx \left(1-\frac{x}{x_{max}}\right) $$ $$      (7.1)
 * $$\displaystyle
 *
 * }

Here, $$\displaystyle x_{max}=15;\ r = 1.4,\ $$ and $$\displaystyle t \in [0,20] $$

The analytical solution is given by the S10 lecture note38-3 as following;

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

$$\displaystyle x(t) = \frac{x_0x_{max}e^{rt}}{x_{max} + x_0(e^{rt}-1)} $$ $$      (7.2)
 * $$\displaystyle
 *
 * }

Logistic equation defined by eqn 7.1 for two initial conditions, $$\displaystyle x_0=3\ and\ x_0 = 9,\ $$, can be solved by using Inconsistent(trapez.) Simpson's rule given by the s10 lecture note 38-3,

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

$$\displaystyle z_{i+1} = z_{i} + \frac{h/2}{3} \left[f_i + 4f_{(i+1/2)} +f_{i+1}\right] $$ $$      (7.3)
 * $$\displaystyle
 *
 * }

The Inconsistent rule (Trapezoidal) is given by,

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

$$\displaystyle z_{i+1/2} = \frac{1}{2} [z_{i} + z_{i+1}] $$      (7.4)
 *
 * }

Solution
By considering the eqn 7.1 in which $$\displaystyle f_{(i+1/2)}$$ is a function of $$\displaystyle z_{i}$$ and $$\displaystyle z_{i+1}$$, where $$\displaystyle z_i, z_{i+1} \rightarrow$$ values of $$\displaystyle x$$ at $$\displaystyle t = t_i$$ and $$\displaystyle t = t_{i+1}$$. This is an initial value problem, $$\displaystyle z_i$$ is known. So, eqn 7.3 as a whole becomes a function of $$\displaystyle z_{i+1}$$ given by,

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

$$\displaystyle F(z_{i+1}) = 0 $$ $$      (7.5) $$\displaystyle z_{i+1}$$ of eqn 7.5 can be found by using Newton-Raphson method
 * $$\displaystyle
 *
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle z_{i+1}^{(k+1)} = z_{i+1}^{(k)} - \left[\frac{dF(z_{i+1}^{(k)})}{dz}\right]^{-1} F(z_{i+1}^{(k)}) $$ $$      (7.6) The Newton-Raphson iteration, starting with an initial guess as $$\displaystyle z_{i+1}^{0} = z_{i}$$, is stopped once the absolute tolerance reaches appropriate criterion.
 * $$\displaystyle
 *
 * }

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

$$\displaystyle \begin{align} AbsTol &= ||z_{i+1}^{(k+1)} - z_{i+1}^{(k)}|| \le 10^{-6} \end{align} $$
 * }
 * }

$$\displaystyle x_0 = 3 $$ :- A comparison on $$\displaystyle h $$ values:

$$\displaystyle x_0 = 9 $$ :- A comparison on $$\displaystyle h $$ values:

=Contributing Members & Referenced Lecture=

title: User:EGM6341.s11.TEAM1.HW7


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


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text: File:Linkerkant.pdf

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text: File:Rechterkant.pdf

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text: File:Boven.pdf

title: File:Boven.pdf


text: File:Onder.pdf

title: File:Onder.pdf


text: File:Aircraft scaledown.svg

title: File:Aircraft scaledown.svg


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User:HK Arun

HK

title: User:HK Arun


text: File:Regimes of gas flow-continuuum to free molecular flow.png


text: File:Logo Helios.pdf

title: File:Logo Helios.pdf


text: User:Andrey Zhivotov

title: User:Andrey Zhivotov


text:
User talk:WouterTh

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User talk:Andrey Zhivotov

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

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --mikeu talk 13:21, 24 March 2011 (UTC)

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User:Eml5526.s11.team2/hwk5

=Problem 5.1: Deriving the Weak Form From the Strong Form (Data 1b)=

Given: Strong Form, Boundary Conditions, and Basis Functions
The Strong Form from lecture 9-1 and Fish and Belytschko p. 73:

Find: The Weak Form in Order to Solve for $$ u^h(x) $$ with Following Basis Functions
Solve for $$ \displaystyle u(x) $$ and $$ \displaystyle u^h(x) $$ using the following basis functions until the error $$ \displaystyle e(n) =|u^h(0.5)-u(0.5)| \leq 10^{-6}$$.

 * 1) Polynomial Basis Function  $$ \displaystyle \mathcal{F}_P :=\{ x^j,j=0,1,2,...\} $$
 * 2) Exponential Basis Function $$ \displaystyle \mathcal{F}_e :=\{ e^{jx},j=0,1,2,...\} $$
 * 3) Fourier Basis Function     $$ \qquad \displaystyle \mathcal{F}_F :=\{cos(jx),j=0,1,2,...,sin(kx),k=1,2,...\} $$

On the domain defined by,

1. Approximating Solution Using Polynomial Basis Function
Here is the matlab code used to obtain the approximating solutions:

Multiplying Equ 2.20 by w(x) and integrating within the domain:

Next we will integrate by parts to obtain,

Since $$ \displaystyle w(x=0) = 0 $$ we can simplify (2.22) in the following way to obtain the weak form,

Now we can use the polynomial basis functions,

Which allows us to write the weak form (2.25) in terms of our matrices $$ \displaystyle {\color{red}\mathbf{K}_{FF}} $$, $$ \displaystyle {\color{red}\mathbf{d}_{F}} $$ and  $$ \displaystyle {\color{red}\mathbf{F}_{F}} $$.

Next we wish to obtain the solution to $$ \displaystyle \mathbf{d_F} $$ by solving the following system of equations,

Solving for the approximating solution, the d matrix can transposed and multiplied by the basis function matrix:

Using the provided matlab code to obtain the d coefficients and applying to the basis function:

n=8
We found that for $$ \displaystyle n=8 $$ the error was of the order $$ \displaystyle 10^{-6} $$. Using the Matlab code and rounding off to the nearest hundredths place gives the matrices $$ \displaystyle \left[ \mathbf{K_{FF}} \right]_ $$ and $$ \displaystyle \left[ \mathbf{F_{F}} \right]_ $$

Using our Matlab code we obtain the following solution for $$ \displaystyle \left[ \mathbf{d_{F}} \right]_ $$. We have chosen to report these values in the long format to demonstrate that rounding error has not occurred.

Plot: Comparing the exact solution $$ u(x) $$ to $$ u_h(x) $$ for $$ n=2,4,6,8 $$
The plots for solutions and error at x=.5 are as follows:

Plot: Showing the error $$ e(n) = |u(x)-u_h(x)| $$
For the polynomial approximate solution, with the increasing number of nodes, the error decreases. For this particular instance, the error XXXXXXX

2. Approximating Soultion Using Fourier Basis Function
Here is the matlab code used to obtain the approximating solutions:

Using the same process as the polynomial solution and using the general equation from equation 2.21, the fourier solution is obtained the same way. W(x) is now equation 5.35 from HW4

Using the provided matlab code to obtain the d coefficients and applying to the basis function:

n=8
Using the Matlab code and rounding off to the nearest hundredths place gives the matrices $$ \displaystyle \left[ \mathbf{K_{FF}} \right]_ $$ and $$ \displaystyle \left[ \mathbf{F_{F}} \right]_ $$

Using our Matlab code we obtain the following solution for $$ \displaystyle \left[ \mathbf{d_{F}} \right]_ $$. We have chosen to report these values in the long format to demonstrate that rounding error has not occurred.

Plot: Comparing the exact solution $$ u(x) $$ to $$ u_h(x) $$ for $$ n=2,4,6,8 $$
The plots for solutions and error at x=.5 are as follows:

3. Approximating Soultion Using Exponential Basis Function
Here is the matlab code used to obtain the approximating solutions:

Using the same process as the polynomial solution and using the general equation from equation 2.21, the exponential solution is obtained the same way. W(x) is now equation 5.41 from HW4

Using the provided matlab code to obtain the d coefficients and applying to the basis function:

n=8
Using the Matlab code and rounding off to the nearest hundredths place gives the matrices $$ \displaystyle \left[ \mathbf{K_{FF}} \right]_ $$ and $$ \displaystyle \left[ \mathbf{F_{F}} \right]_ $$

Using our Matlab code we obtain the following solution for $$ \displaystyle \left[ \mathbf{d_{F}} \right]_ $$. We have chosen to report these values in the long format to demonstrate that rounding error has not occurred.

Plots
The plots for solutions and error at x=.5 are as follows:

=Problem 5.2: Deriving the Weak Form From the Strong Form (Data 1)=

Given: Strong Form, Boundary Conditions, and Basis Functions
The Strong Form from lecture 9-1 and Fish and Belytschko p. 73:

Find: The Weak Form in Order to Solve for $$ u^h(x) $$ with Following Basis Functions
Solve for $$ \displaystyle u(x) $$ and $$ \displaystyle u^h(x) $$ using the following basis functions until the error $$ \displaystyle e(n) =|u^h(0.5)-u(0.5)| \leq 10^{-6}$$.

 * 1) Polynomial Basis Function  $$ \displaystyle \mathcal{F}_P :=\{ x^j,j=0,1,2,...\} $$
 * 2) Exponential Basis Function $$ \displaystyle \mathcal{F}_e :=\{ e^{jx},j=0,1,2,...\} $$
 * 3) Fourier Basis Function     $$ \qquad \displaystyle \mathcal{F}_F :=\{cos(jx),j=0,1,2,...,sin(kx),k=1,2,...\} $$

On the domain defined by,

1. Approximating Solution Using Polynomial Basis Function
Here is the matlab code used to obtain the approximating solutions:

Multiplying Equ 2.20 by w(x) and integrating within the domain:

Next we will integrate by parts to obtain,

Since $$ \displaystyle w(x=1) = 0 $$ we can simplify (2.22) in the following way to obtain the weak form,

Now we can use the polynomial basis functions,

Which allows us to write the weak form (2.25) in terms of our matrices $$ \displaystyle {\color{red}\mathbf{K}_{FF}} $$, $$ \displaystyle {\color{red}\mathbf{d}_{F}} $$ and  $$ \displaystyle {\color{red}\mathbf{F}_{F}} $$.

Next we wish to obtain the solution to $$ \displaystyle \mathbf{d_F} $$ by solving the following system of equations,

Solving for the approximating solution, the d matrix can transposed and multiplied by the basis function matrix:

Using the provided matlab code to obtain the d coefficients and applying to the basis function:

n=8
We found that for $$ \displaystyle n=8 $$ the error was of the order $$ \displaystyle 10^{-6} $$. Using the Matlab code and rounding off to the nearest hundredths place gives the matrices $$ \displaystyle \left[ \mathbf{K_{FF}} \right]_ $$ and $$ \displaystyle \left[ \mathbf{F_{F}} \right]_ $$

Using our Matlab code we obtain the following solution for $$ \displaystyle \left[ \mathbf{d_{F}} \right]_ $$. We have chosen to report these values in the long format to demonstrate that rounding error has not occurred.

Plot: Comparing the exact solution $$ u(x) $$ to $$ u_h(x) $$ for $$ n=2,4,6,8 $$
The plots for solutions and error at x=.5 are as follows:

Plot: Showing the error $$ e(n) = |u(x)-u_h(x)| $$
For the polynomial approximate solution, with the increasing number of nodes, the error decreases. For this particular instance, the error XXXXXXX

2. Approximating Soultion Using Fourier Basis Function
Here is the matlab code used to obtain the approximating solutions:

Using the same process as the polynomial solution and using the general equation from equation 2.21, the fourier solution is obtained the same way. W(x) is now equation 5.35 from HW4

Using the provided matlab code to obtain the d coefficients and applying to the basis function:

n=4
etc...

n=8
Using the Matlab code and rounding off to the nearest hundredths place gives the matrices $$ \displaystyle \left[ \mathbf{K_{FF}} \right]_ $$ and $$ \displaystyle \left[ \mathbf{F_{F}} \right]_ $$

Plot: Comparing the exact solution $$ u(x) $$ to $$ u_h(x) $$ for $$ n=2,4,6,8 $$
The plots for solutions and error at x=.5 are as follows:

3. Approximating Soultion Using Exponential Basis Function
Here is the matlab code used to obtain the approximating solutions:

Using the same process as the polynomial solution and using the general equation from equation 2.21, the exponential solution is obtained the same way. W(x) is now equation 5.41 from HW4

Using the provided matlab code to obtain the d coefficients and applying to the basis function:

n=8
Using the Matlab code and rounding off to the nearest hundredths place gives the matrices $$ \displaystyle \left[ \mathbf{K_{FF}} \right]_ $$ and $$ \displaystyle \left[ \mathbf{F_{F}} \right]_ $$

Using our Matlab code we obtain the following solution for $$ \displaystyle \left[ \mathbf{d_{F}} \right]_ $$. We have chosen to report these values in the long format to demonstrate that rounding error has not occurred.

The plots for solutions and error at x=.5 are as follows:

Matlab
=Problem 5.3: Torsion of a Bar = Modified version of HW 4.4 defined on page 2 of Lecture 21 from Fish & Belytschko Page 73 Problem 3.7

Given
Where the Weak Form for a bar in torsion can be defined as

Make use of the Lagrange Interpolation Basis Function (LIBF) with uniform discretization (equidistant nodes) using $$ \displaystyle m = 4,6,8   $$ where the LIBF can defined as follows

Find
a) Explain how Langrange Interpolation Basis Functions (LIBF) are used as Constraint Breaking Solutions

b) Plot all LIBF used

c) Use matlab quad function or WolframAlpha.com to integrate

d) Plot approximate vs exact solution and convergence error versus equidistant nodes (m)

(a)
Langrange Interpolation Basis Functions satisfy the condition of the constraint breaking solution by assuming a value of zero at every node except for the node where the LIBF is defined there it assumes a value of 1. It acts like the Kronecker Delta

(b)
The following is the LIBF for the $$ \displaystyle m = 4 $$

The following is the LIBF for the $$ \displaystyle m = 6 $$

The following is the LIBF for the $$ \displaystyle m = 8 $$

Plot: Comparing the exact solution $$ u(x) $$ to $$ u_h(x) $$ for $$ n=2,4,6,8 $$
The plots for solutions and error at x=.5 are as follows:

Matlab Code
=Problem 5.4: Torsion of a Bar = Modified version of HW 4.4 defined on page 2 of Lecture 21 from Fish & Belytschko Page 73 Problem 3.7

Given
Where the Weak Form for a bar in torsion can be defined as

Make use of the Lagrange Interpolation Basis Function (LIBF) with uniform discretization (equidistant nodes) using $$ \displaystyle m = 4,6,8   $$ where the LIBF can defined as follows

Find
a) Explain how Langrange Interpolation Basis Functions (LIBF) are used as Constraint Breaking Solutions b) Plot all LIBF used c) Use matlab quad function or WolframAlpha.com to integrate d) Plot approximate vs exact solution and convergence error versus equidistant nodes (m)

(a)
Langrange Interpolation Basis Functions satisfy the condition of the constraint breaking solution by assuming a value of zero at every node except for the node where the LIBF is defined there it assumes a value of 1. It acts like the Kronecker Delta

(b)
The following is the LIBF for the $$ \displaystyle m = 4 $$

The following is the LIBF for the $$ \displaystyle m = 6 $$

The following is the LIBF for the $$ \displaystyle m = 8 $$

(c)
Our solution $$ u \left ( x \right ) $$ and our weight functions can be approximated using the following definitions,

To approximate our solution u, we divide our x domain into m nodal subdivisions. Additionally, we will use m LIBF to approximate our solution. These LIBF will have the following form,

Therefore, for m=4, we take the first 4 Lagrange Interpolation Basis Functions, or more specifically,

Here we want to develop equations for $$ \tilde{K_{KK}} $$ and $$ \tilde{F_K} $$ and solve for $$ \tilde{d_{K}} $$, similar to the previous problems.

Plot: Comparing the exact solution $$ u(x) $$ to $$ u_h(x) $$ for $$ n=2,4,6,8 $$
The plots for solutions and error at x=.5 are as follows:

Matlab Code
= Problem 5.5 : Calculix-Using CCX module=

Disk Problem
After building disc in cgx module then we might want to know that properties of nodes,lines, surfaces, and bodies so now lets look at how we can get more information what we have done for our model.

First we have to open our program. To do this run the following ;

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$$  \displaystyle [All Programs]\Rightarrow [bConverged]\Rightarrow [Calculix]\Rightarrow [Calculix Command] $$
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 * style="width:95%" |
 * }

Open calculix and go to directory where the basic examples stays in your harddrive.
 * {| style="width:100%" border="0"

$$  \displaystyle ...\Rightarrow [Program Files]\Rightarrow [bConverged] \Rightarrow [Calculix] \Rightarrow [Cgx] \Rightarrow [Examples] \Rightarrow [Basic] $$
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 * }

Then type to open disc example:

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$$  \displaystyle [cgx]\Rightarrow [disc] $$
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 * }

Now we want to learn about point properties. To see point properties in our graph secren type

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$$  \displaystyle \begin{matrix} plot & pa   & all \end{matrix} $$
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 * }

We need to mesh disc to see the mesh properties to mesh the disc please type following steps

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$$  \displaystyle \begin{matrix} elty & all & qu8 \end{matrix} $$
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 * }

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$$  \displaystyle \begin{matrix} mesh & all \end{matrix} $$
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 * }

What we have done so far is that we opened disc and meshed it with quadratic elements. So we are ready to see element properties on our graph screen. To see properties of elements type,
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{matrix} plot & ma  & all \end{matrix} $$
 * style="width:95%" |
 * style="width:95%" |

 * }

Also we can get more specific properties form the special editor of calculix called 'SciTE.' We can go to examples directory and just basically right click on disc example and open with by editing. SciTE automatically will be opened.

From below we can easly see the properties of all components of disc. Especially point properties are boxed.

Generating Different Meshes
We will do the same procedures as we have done before. Our model is prepared. So we will mesh our model and take properties of different kinds of meshes for 2-D. Our meshes for 2-D are:

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

$$  \displaystyle \begin{matrix} tr3 & 3 node & 2D element \\ gu4 & 4 node & 2D element\\ qu8 & 8 node & 2D element \end{matrix} $$     (N)
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 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{matrix} Mesh & with  & tr3 \end{matrix} $$
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 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{matrix} Mesh & with  & qu4 \end{matrix} $$
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 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{matrix} Mesh & with  & qu8 \end{matrix} $$
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 * }

CCX module
In this module we will analyze for unknowns that we are interested in. This unknowns can be displacements or temperatures. Also we can find fluxes between these nodes.

For the first step go to directory where beam example stays in your harddrive.

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$$  \displaystyle ....[Program Files]\Rightarrow [bConverged]\Rightarrow [Calculix]\Rightarrow [ccx]\Rightarrow [test] $$
 * style="width:95%" |
 * style="width:95%" |

 * }

Find beampl.inp and right click on it. Then edit it by selecting open with. SciTe automatically will be opened. Every command inside is ready so we have assumed the we have completed by the mesh and ready for solving.So click tools and click 'solve'.

You have to see that $$ \displaystyle Job Finished $$  Then click tools and click 'post prosess'.

Results
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$$  \displaystyle Result 1 $$
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 * }
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$$  \displaystyle Result2 $$
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 * }

=Problem 5.6: The Quadratic Lagrangian Element Basis Function = Lecture 30-6

Given:
Given the Linear Lagrangian Element Basis Function,

from Lecture 30-6

Find:
The Quadratic Lagrangian Element Basis Function

Solution:
The above is the depiction of the basis function satisfying the CBS for element node i-1 The above is the depiction of the basis function satisfying the CBS for element node i The above is the depiction of the basis function satisfying the CBS for element node i+1 Next we overlay the three functions to show the shape function for the quadratic Lagrangian

=Problem 5.7 Weak form solution using LLEIF=

Given
Solve the following discrete weak form with Linear Lagrange Element Interpolation Basis.

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$$  \displaystyle \sum {{c_i}\left[ {\sum {{{\tilde K}_{ij}}{d_j} - {{\tilde F}_i}} } \right]} = 0 $$     (3.7.1)
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 *
 * }

Where

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$$  \displaystyle {{\tilde K}_{ij}} = \int\limits_0^1 {{{b'}_i}(2 + 3x){{b'}_j}dx} $$     (3.7.2)
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 * style="width:95%" |
 *
 * }
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$$  \displaystyle {F_i} = {b_i}\left( 1 \right)\left( {12} \right) + \int_0^1 {{b_i}\left( x \right)5x} dx $$ (3.7.3)
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 *
 * }

Such that

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

$$  \displaystyle {w^h}\left( 0 \right) = \sum {{c_i}{b_i}\left( 0 \right) = 0} $$     (3.7.5)
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 *
 * }

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

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

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

$$  \displaystyle {w^h}\left( 0 \right) = \sum {{c_i}{b_i}\left( 0 \right) = 0} $$     (3.7.6)
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 *
 * }

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

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

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

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

$$  \displaystyle {w^h} = {c_1}(1) + {c_2}(0) + {c_3}(0) + {c_4}(0) + \cdots $$     (3.7.8)
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 *
 * }

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$$  \displaystyle {w^h} = {c_1} $$     (3.7.9)
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 *
 * }

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

$$  \displaystyle {w^h}\left( 0 \right) = 0 $$     (3.7.10)
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 *
 * }

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$$  \displaystyle {c_1} = 0 $$     (3.7.11)
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 *
 * }

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

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

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$$  \displaystyle \Omega = \bigcup\limits_{e = 1}^{nel} $$     (3.7.12)
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 *
 * }

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

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

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

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

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

$$  \displaystyle {{\tilde F}_i} = \sum\limits_{e = 1}^{nel} {F_i^e} $$     (3.7.15)
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 *
 * }

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

$$  \displaystyle {F_i}^e = \int_ {b_i^e5xdx} $$     (3.7.16)
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 * <p style="text-align:right">
 * }

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

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$$  \displaystyle {b_e} = \frac $$     (3.7.17)
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 * <p style="text-align:right">
 * }

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

$$  \displaystyle {b_{e + 1}} = \frac $$     (3.7.18)
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 * <p style="text-align:right">
 * }

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

Plot of exact solution and approximate solution with 4 elements

Plot of exact solution and approximate solution with 6 elements

Plot of exact solution and approximate solution with 8 elements

Plot of error vs dof's

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

$$  \displaystyle err = u\left( {0.5} \right) - {u^h}\left( {0.5} \right) $$     (3.7.19)
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 * <p style="text-align:right">
 * }

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

$$  \displaystyle dof's = nel + 1 $$     (3.7.20)
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 * <p style="text-align:right">
 * }

where nel= number of elements.

for 16 elements

dof's=16+1=17

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

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

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

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

$$  \displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\left [ (2+3x)\frac{\mathrm{d} u}{\mathrm{d} x} \right ]+5x=0 $$     (8.1)
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 * <p style="text-align:right">
 * }

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

$$  \displaystyle u(1)=4 $$     (8.2)
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 * <p style="text-align:right">
 * }

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$$  \displaystyle \frac{\mathrm{d} u(x=0)}{\mathrm{d} x}=-3 $$     (8.3)
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 * <p style="text-align:right">
 * }

Weak Form
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$$  \displaystyle \int_{\Omega }\frac{\mathrm{d} w}{\mathrm{d} x}AE\frac{\mathrm{d} u}{\mathrm{d} x}dx-\left ( wAt \right )_{\Gamma _{h}}-\int_{\Omega }wbdx=0 \Rightarrow \Omega =]0,1[ $$     (8.4)
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 * <p style="text-align:right">
 * }

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

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

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

Approximate functions gives as in F&B Book,

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$$  \displaystyle u^{h}=\sum_{e=1}^{nel}N^{e}d^{e}=\left ( \sum_{e=1}^{nel}N^{e}L^{e}\right )\mathbf{d} $$     (8.6)
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 * <p style="text-align:right">
 * }

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

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

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

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

$$  \displaystyle N^{e}=\frac{1}{l^{e}}[\begin{matrix} x_{2}^{e}-x & x-x_{1}^{e} \end{matrix}] $$     (8.8)
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 * }

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$$  \displaystyle B^{e}=\frac{1}{l^{e}}[\begin{matrix} -1 & 1 \end{matrix}] $$     (8.9)
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 * <p style="text-align:right">
 * }

Analytical Solution

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

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

$$  \displaystyle 2\frac{\mathrm{d} u}{\mathrm{d} x}(x=0)=0+c_{1} \Rightarrow c_{1}=-6 $$     (8.11)
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 * }
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$$  \displaystyle u=-\frac{5}{2}\int \frac{x^2}{2+3x}dx-\int \frac{6}{2+3x}dx+c_{2} $$     (8.12)
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$$  \displaystyle -\frac{5}{2}\left [ \frac{1}{54}(9x^2-12x+8log(3x+2)-12) \right ]-\left [ \frac{1}{3}log(3x+2) \right ]+c_{2} $$     (8.13)
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Using other boundry condition ;

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$$  \displaystyle c_{2}=\frac{119}{36}+\frac{64}{27}log(5) $$     (8.14)
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$$  \displaystyle u(x)=-\frac{5}{12}x^2+\frac{5}{9}x-\frac{64}{27}log(2+3x)+\frac{139}{36}+\frac{64}{27}log(5) $$     (8.15)
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FEA solutions
When evaluating unknowns on distributed nodes our shape functions must meet some conditions.Our approximate functions must be complete in each element. As saying completeness we are saying that by using our nodal values we must be able to identify coefficients in our approximate solutions. After some manipulation we conclude with shape functions. For each nodes we have shape functions and for middle nodes shape functions come from adjacent elements.

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

$$  \displaystyle N=N^{(1)}L^{(1)}+N^{(2)}L^{(2)}=[\begin{matrix} N_{1}^{(1)} & N_{2}^{(1)}+N_{1}^{(2)} & N_{2}^{(2)} \end{matrix}]
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$$     (8.16)
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For this example at x1 (first node) only N1 will be 1. Others will be zero. So we verify CBS.

Using 4 elements

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$$  \displaystyle \Omega =]0,1[ $$ [[image:Egm6321.f10.team2.oztekin.Elemtdiv.png|center]]
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 * }

Lets find shape functions for 4 elements.

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$$  \displaystyle\begin{align} &N^{(1)}=\frac{1}{0.25}[\begin{matrix} 0.25-x & x \end{matrix}]\\ &N^{(2)}=\frac{1}{0.25}[\begin{matrix} 0.5-x & x-0.25 \end{matrix}]\\ &N^{(3)}=\frac{1}{0.25}[\begin{matrix} 0.75-x & x-0.5 \end{matrix}]\\ &N^{(4)}=\frac{1}{0.25}[\begin{matrix} 1-x & x-0.75 \end{matrix}] \end{align} $$     (8.17)
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These shape functions are useful for us to make connection between nodes. After defining these tools now we can find stiffness matrices for all elements. Then after finding all of stiffness matrices for elements we can construct global stiffness matrix.

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$$  \displaystyle K^{e}=\int_{x_{1}^{e}}^{x_{2}^{e}}B^{eT}A^{e}E^{e}B^{e}dx $$     (8.18)
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$$  \displaystyle\begin{align} {{K}^{e}}&=\int\limits_{x_{1}^{e}}^{x_{2}^{e}}{\frac{1}\left[ \begin{matrix} -1 \\   1  \\ \end{matrix} \right]}(2+3x)\frac{1}[\begin{matrix} -1 & 1 \\ \end{matrix}]dx \\ &=\frac{1}\left[ \begin{matrix} 1 & -1 \\   -1 & 1  \\ \end{matrix} \right]\left[ 2(x_{2}^{e}-x_{1}^{e})+\frac{3}{2}(x_{2}^{e2}-x_{1}^{e2}) \right] \end{align} $$     (8.19)
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 * }

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

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$$  \displaystyle K^{(1)}=\begin{bmatrix} 9.5 & -9.5\\ -9.5& 9.5 \end{bmatrix} $$     (8.21)
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 * }

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$$  \displaystyle K^{(2)}=\begin{bmatrix} 12.5 & -12.5\\ -12.5& 12.5 \end{bmatrix} $$     (8.22)
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 * }

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

$$  \displaystyle K^{(3)}=\begin{bmatrix} 15.5 & -15.5\\ -15.5& 15.5 \end{bmatrix} $$     (8.23)
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 * }

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$$  \displaystyle K^{(4)}=\begin{bmatrix} 18.5 & -18.5\\ -18.5& 18.5 \end{bmatrix} $$     (8.24)
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 * }

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

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$$  \displaystyle K=\sum_{e=1}^{nel}L^{eT}K^{e}L^{e}=L^{1T}K^{1}L^{1}+L^{2T}K^{2}L^{2}+L^{3T}K^{3}L^{3}+L^{4T}K^{4}L^{4} $$     (8.25)
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 * }
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$$  \displaystyle K=\left[ \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} 9.5 \\   -9.5  \\ \end{matrix}  \\ 0 \\   0  \\   0  \\ \end{matrix} & \begin{matrix} \begin{matrix} -9.5 \\   22  \\ \end{matrix}  \\ -12.5 \\   0  \\   0  \\ \end{matrix}  \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 \\   -12.5  \\ \end{matrix}  \\ 28 \\   \begin{matrix} -15.5 \\   0  \\ \end{matrix}  \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 \\   0  \\ \end{matrix}  \\ -15.5 \\   34  \\   -18.5  \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 \\   0  \\ \end{matrix}  \\ 0 \\   -18.5  \\   18.5  \\ \end{matrix}  \\ \end{matrix} \right]
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$$     (8.26)
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 * }

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

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

$$  \displaystyle f_{\Gamma _{h}}=-(N^{eT}A^{e}b)_{\Gamma _{h}^{e}}=-N^{eT}(x_{1})(2+3x)_{x=0}(-3)=6N^{eT}(x_{1}) $$     (8.27)
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$$  \displaystyle f_{{\Gamma }^{(1)}}=6\left[ \begin{matrix} N_{1}^{(1)}({{x}_{1}}) \\ N_{2}^{(1)}({{x}_{2}}) \\ \end{matrix} \right]=\left[ \begin{matrix} 6 \\   0  \\ \end{matrix} \right]
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$$     (8.28)
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 * }

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

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$$  \displaystyle\begin{align} f_{\Gamma _{h}} &=\sum_{e=1}^{nel}L^{eT}f_{\Gamma _{h}}^{e} \\ &=\left[ \begin{matrix} \begin{matrix} 6 \\   0  \\ \end{matrix}  \\ 0 \\   0  \\   0  \\ \end{matrix} \right]
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\end{align} $$     (N)
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 * }

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$$  \displaystyle\begin{align} f_{\Omega }^{e} & =\int_{x_{1}^{e}}^{x_{2}^{e}}N^{eT}b(x)dx\\ &=\int\limits_{x_{1}^{e}}^{x_{2}^{e}}{\frac{1}\left[ \begin{matrix} x_{2}^{e}-x \\ x-x_{1}^{e} \\ \end{matrix} \right]}5xdx \\ &=\frac{5}\left[ \begin{matrix} \frac{x_{2}^{e}(x_{2}^{e2}-x_{1}^{e2})}{2}-\frac{(x_{2}^{e3}-x_{1}^{e3})}{3} \\ \frac{(x_{2}^{e3}-x_{1}^{e3})}{3}-\frac{x_{1}^{e}(x_{2}^{e2}-x_{1}^{e2})}{2} \\ \end{matrix} \right]
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\end{align} $$     (8.30)
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 * }

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$$  \displaystyle f_{\Omega }^{(1)}=\left[ \begin{matrix} 0.05208334 \\   0.10416666  \\ \end{matrix} \right] $$     (8.31)
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 * }

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$$  \displaystyle f_{\Omega }^{(2)}=\left[ \begin{matrix} 0.2083334 \\   0.2604166  \\ \end{matrix} \right] $$     (8.32)
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 * }

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$$  \displaystyle f_{\Omega }^{(3)}=\left[ \begin{matrix} 0.36458334 \\   0.41666666  \\ \end{matrix} \right] $$     (8.33)
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 * }

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$$  \displaystyle f_{\Omega }^{(4)}=\left[ \begin{matrix} 0.52083334 \\   0.57291666  \\ \end{matrix} \right] $$     (8.34)
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 * }

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$$  \displaystyle\begin{align} f_{\Omega }&=\sum_{e=1}^{4}L^{eT}f_{\Omega }^{e} &\left[ \begin{matrix} \begin{matrix} 0.05208334 \\   0.3125  \\ \end{matrix}  \\ 0.625 \\   0.9375  \\   0.57291666  \\ \end{matrix} \right]
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\end{align} $$     (8.35)
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$$  \displaystyle \left[ \begin{matrix} \begin{matrix} 9.5 \\   -9.5  \\ \end{matrix} & \begin{matrix} -9.5 \\   22  \\ \end{matrix} & \begin{matrix} 0 \\   -12.5  \\ \end{matrix} & \begin{matrix} \begin{matrix} 0 & 0 \\ \end{matrix}  \\ \begin{matrix} 0 & 0 \\ \end{matrix}  \\ \end{matrix} \\ 0 & -12.5 & 28 & \begin{matrix} -15.5 & 0 \\ \end{matrix}  \\ 0 & 0 & -15.5 & \begin{matrix} 34 & -18.5 \\ \end{matrix}  \\ 0 & 0 & 0 & \begin{matrix} -18.5 & 18.5 \\ \end{matrix}  \\ \end{matrix} \right]\left[ \begin{matrix} \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ \end{matrix} \\ {{d}_{3}} \\ {{d}_{4}} \\ 4 \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} 0.05208334 \\   0.3125  \\ \end{matrix}  \\ 0.625 \\   0.9375  \\   0.57291666  \\ \end{matrix} \right]+\left[ \begin{matrix} \begin{matrix} 6 \\   0  \\ \end{matrix}  \\ 0 \\   0  \\   0  \\ \end{matrix} \right]+\left[ \begin{matrix} \begin{matrix} 0 \\   0  \\ \end{matrix}  \\ 0 \\   0  \\   {{r}_{1}}  \\ \end{matrix} \right]
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$$     (8.36)
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$$  \displaystyle \left[ \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ {{d}_{3}} \\ {{d}_{4}} \\ \end{matrix} \right]=\left[ \begin{matrix} 2.0257 \\   1.3886  \\   0.8794  \\   0.4285  \\ \end{matrix} \right]
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$$     (8.37)
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Comparing results with exact ones at the nodes;

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$$  \displaystyle\begin{align} &\left[ \begin{matrix} 2.0257 \\   1.3886  \\   0.8794  \\   0.4285  \\ \end{matrix} \right]+\left[ \begin{matrix} 4 \\   4  \\   4  \\   4  \\ \end{matrix} \right]=\left[ \begin{matrix} 6.0257 \\   5.3886  \\   4.8794  \\   4.4285  \\ \end{matrix} \right]
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&{{d}_{exact}}=\left[ \begin{matrix} 6.0331 \\   5.3911  \\   4.8802  \\   4.4286  \\ \end{matrix} \right]

\end{align} $$     (8.38)
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$$  \displaystyle 4 elements $$
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Using 6 elements
Shape functions for 6 elements
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$$  \displaystyle\begin{align} & N^{(1)}=\frac{1}{0.1666}[\begin{matrix} 0.1666-x & x \end{matrix}]\\ &N^{(2)}=\frac{1}{0.1666}[\begin{matrix} 0.3333-x & x-0.16666 \end{matrix}]\\ & N^{(3)}=\frac{1}{0.1666}[\begin{matrix} 0.5-x & x-0.3333 \end{matrix}]\\ &N^{(4)}=\frac{1}{0.1666}[\begin{matrix} 0.6666-x & x-0.5 \end{matrix}]\\ & N^{(5)}=\frac{1}{0.1666}[\begin{matrix} 0.8332-x & x-0.6666 \end{matrix}]\\ & N^{(6)}=\frac{1}{0.1666}[\begin{matrix} 1-x & x-0.8332 \end{matrix}]\\
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\end{align} $$     (8.39)
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$$  \displaystyle \left[ \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} 13.5048 & -13.5048 & 0 & 0 \\ \end{matrix} & 0 & 0 & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} -13.5048 & 30.0105 & -16.5057 & 0 \\ \end{matrix} & 0 & 0 & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & -16.5057 & 36.0132 & -19.5075 \\ \end{matrix} & 0 & 0 & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 & -19.5075 & 42.0157 \\ \end{matrix} & -22.5084 & 0 & 0  \\ \end{matrix} \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 & 0 & -22.5084 \\ \end{matrix} & 48.0164 & -25.508 & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 & 0 & 0 \\ \end{matrix} & -25.508 & 54.008 & -28.5  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 & 0 & 0 \\ \end{matrix} & 0 & -28.5 & 28.5  \\ \end{matrix} \\ \end{matrix} \right]\left[ \begin{matrix} \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ {{d}_{3}} \\ {{d}_{4}} \\ \end{matrix} \\ {{d}_{5}} \\ {{d}_{6}} \\ 4 \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} 6.0232199942 \\   0.13890282  \\   0.277972087  \\   0.416641681  \\ \end{matrix}  \\ 0.555277743 \\   0.694970317  \\   0.394287575  \\ \end{matrix} \right]+\left[ \begin{matrix} \begin{matrix} 0 \\   0  \\   0  \\   0  \\ \end{matrix}  \\ 0 \\   0  \\   {{r}_{1}}  \\ \end{matrix} \right]
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$$     (8.40)
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$$  \displaystyle \left[ \begin{matrix} \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ {{d}_{3}} \\ \end{matrix} \\ {{d}_{4}} \\ {{d}_{5}} \\ {{d}_{6}} \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} 2.0292 \\   1.5831  \\   1.2098  \\ \end{matrix}  \\ 0.8797 \\   0.575  \\   0.2845  \\ \end{matrix} \right]
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$$     (8.41)
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$$  \displaystyle\begin{align} & \left[ \begin{matrix} \begin{matrix} 2.0292 \\   1.5831  \\   1.2098  \\ \end{matrix}  \\ 0.8797 \\   0.575  \\   0.2845  \\ \end{matrix} \right]+\left[ \begin{matrix} \begin{matrix} 4 \\   4  \\   4  \\ \end{matrix}  \\ 4 \\   4  \\   4  \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} 6.0292 \\   5.5831  \\   5.2098  \\ \end{matrix}  \\ 4.8797 \\   4.575  \\   4.2845  \\ \end{matrix} \right]
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& {{d}_{exact}}=\left[ \begin{matrix} \begin{matrix} 6.0331 \\   5.5853  \\   5.2109  \\ \end{matrix}  \\ 4.8802 \\   4.5753  \\   4.2847  \\ \end{matrix} \right]

\end{align} $$     (8.42)
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$$  \displaystyle 6 elements $$
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 * }

Using 8 elements
Shape functions for 8 elements
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$$  \displaystyle\begin{align} & N^{(1)}=\frac{1}{0.125}[\begin{matrix} 0.125-x & x \end{matrix}]\\ &N^{(2)}=\frac{1}{0.125}[\begin{matrix} 0.25-x & x-0.125 \end{matrix}]\\ & N^{(3)}=\frac{1}{0.125}[\begin{matrix} 0.375-x & x-0.25 \end{matrix}]\\ &N^{(4)}=\frac{1}{0.125}[\begin{matrix} 0.5-x & x-0.375 \end{matrix}]\\ & N^{(5)}=\frac{1}{0.125}[\begin{matrix} 0.625-x & x-0.5 \end{matrix}]\\ & N^{(6)}=\frac{1}{0.125}[\begin{matrix} 0.75-x & x-0.625 \end{matrix}]\\ & N^{(7)}=\frac{1}{0.125}[\begin{matrix} 0.875-x & x-0.75 \end{matrix}]\\ & N^{(8)}=\frac{1}{0.125}[\begin{matrix} 1-x & x-0.875 \end{matrix}]\\ \end{align} $$     (8.43)
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 * }

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$$  \displaystyle \left[ \begin{matrix} 17.5 & -17.5 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\   -17.5 & 38 & -20.5 & 0 & 0 & 0 & 0 & 0 & 0  \\   0 & -20.5 & 44 & -23.5 & 0 & 0 & 0 & 0 & 0  \\   0 & 0 & -23.5 & 50 & -26.5 & 0 & 0 & 0 & 0  \\   0 & 0 & 0 & -26.5 & 56 & -29.5 & 0 & 0 & 0  \\   0 & 0 & 0 & 0 & -29.5 & 62 & -32.5 & 0 & 0  \\   0 & 0 & 0 & 0 & 0 & -32.5 & 68 & -35.5 & 0  \\   0 & 0 & 0 & 0 & 0 & 0 & -35.5 & 74 & -38.5  \\   0 & 0 & 0 & 0 & 0 & 0 & 0 & -38.5 & 38.5  \\ \end{matrix} \right]\left[ \begin{matrix} \begin{matrix} \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ {{d}_{3}} \\ \end{matrix} \\ {{d}_{4}} \\ {{d}_{5}} \\ {{d}_{6}} \\ \end{matrix} \\ {{d}_{7}} \\ {{d}_{8}} \\ {{d}_{9}} \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} 6.01302084 \\   0.078125  \\ \end{matrix}  \\ 0.15625 \\ \end{matrix}  \\ 0.234375 \\   0.31250004  \\   0.390625  \\ \end{matrix}  \\ 0.46874373 \\   0.546875  \\   0.29947916  \\ \end{matrix} \right]+\left[ \begin{matrix} \begin{matrix} \begin{matrix} 0 \\   0  \\   0  \\ \end{matrix}  \\ 0 \\   0  \\   0  \\ \end{matrix}  \\ 0 \\   0  \\   {{r}_{1}}  \\ \end{matrix} \right]
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$$     (8.44)
 * <p style="text-align:right">
 * }

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

$$  \displaystyle \left[ \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} {{d}_{1}} \\ {{d}_{2}} \\ \end{matrix} \\ {{d}_{3}} \\ \end{matrix} \\ {{d}_{4}} \\ {{d}_{5}} \\ \end{matrix} \\ {{d}_{6}} \\ {{d}_{7}} \\ {{d}_{8}} \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} \begin{matrix} 2.0312 \\   1.6876  \\ \end{matrix}  \\ 1.3904 \\   1.1246  \\   0.8800  \\ \end{matrix}  \\ 0.6497 \\   0.4286  \\   0.213  \\ \end{matrix} \right]
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$$     (8.45)
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 * }

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

$$  \displaystyle\begin{align} &\left[ \begin{matrix} \begin{matrix} \begin{matrix} 2.0312 \\   1.6876  \\ \end{matrix}  \\ 1.3904 \\   1.1246  \\   0.8800  \\ \end{matrix}  \\ 0.6497 \\   0.4286  \\   0.213  \\ \end{matrix} \right]+\left[ \begin{matrix} \begin{matrix} \begin{matrix} 4 \\   4  \\ \end{matrix}  \\ 4 \\   4  \\   4  \\ \end{matrix}  \\ 4 \\   4  \\   4  \\ \end{matrix} \right]=\left[ \begin{matrix} \begin{matrix} \begin{matrix} 6.0312 \\   5.6876  \\ \end{matrix}  \\ 5.3904 \\   5.1246  \\   4.88  \\ \end{matrix}  \\ 4.6497 \\   4.4286  \\   4.213  \\ \end{matrix} \right]
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&{{d}_{exact}}=\left[ \begin{matrix} \begin{matrix} \begin{matrix} 6.0331 \\   5.6866  \\ \end{matrix}  \\ 5.3911 \\   5.1249  \\   4.8802  \\ \end{matrix}  \\ 4.6498 \\   4.4286  \\   4.213  \\ \end{matrix} \right]

\end{align} $$     (8.46)
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 * }

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

$$  \displaystyle 8 elements $$
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 * }

=References=

=Contributing Members & Referenced Lecture=

title: User:Eml5526.s11.team2/hwk5


text:
User:Nobody60

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 * 2) Happiness research
 * 3) Positive thinking
 * 4) Everyone needs a spiritual centering
 * 5) The Five Principles of Health
 * 6) Have a Happy Relationship
 * 7) Have a Happy Family
 * 8) Be a Happy Single
 * 9) Have a Happy Old Age
 * 10) Have a Happy Job
 * 11) Happiness and Life
 * 12) Happiness and Enlightenment
 * 13) Laughter Yoga
 * 14) A World of Peace, Love and Happiness
 * 15) Happiness oracle (Tarot)
 * 16) Optimism

title: User:Nobody60


text:
Preventing email spam

If you publish an email address on a website it can open up the possibility of E-mail address harvesting. There are a number of techniques for publishing an email link on the web that can make it more difficult for spammers to abuse the email link. On a MediaWiki website, such as Wikiversity, the easiest method is to use the  feature. For example, to include a link to the email of User:Example you would use the wiki link Special:EmailUser/Example. This assumes that User:Example has entered a valid email in.

Another option is to use a tool such as Obfuscate Email Address on Web Page. If we enter the example email address  it returns code that can be inserted in any webpage, including a wiki. It looks like  in the webpage but if you click on edit this page you will see that the plain text of the email address is obfuscated using html markup and appears as   in the source code. Tbis is not guaranteed to work but could provide some measure of protection against address harvesting.

See the links below for a more detailed discussion of different methods to hide email addresses from spammers.

title: Preventing email spam


text:
Happiness/Philosophy of Happiness

School:Philosophy

Socrates
Socrates (469 BC in Athens – 399 BC) is fundamental for Western thinking. Almost all the major philosophical schools of antiquity have to rely on Socrates. Michel de Montaigne called him the "master of masters" and Karl Jaspers wrote, "Socrates to have in mind is one of the essential conditions of our philosophy". (Eva-Maria Kaufmann: Sokrates. Munich 2000, p. 93 (Montaigne), p. 8 (Jaspers).)

Socrates was a mystic. There are many accounts of his extraordinary abilities. He could fall for hours in meditation, go barefoot in winter and consume excessive amounts of alcohol, without ever showing signs of intoxication. He lived in strict guidance by his inner voice, whose origin he regarded as divine. He prayed regularly. About life after death, he preferred an open perspective, "Either it is a non-being, and we have no sensation after death - or, as it is told, it is a migration of the soul from this place to another."(Plato, Apologie, Stuttgart 1982, p. 36) It is interesting to note that a significant portion of Plato's Pheado is concerned with various arguments whose purpose can be seen to prove the existence of life after death.(Plato, Phaedo, 66a-67d.)

We can regard Socrates as the spiritual father of today's scientific thinking time. He was an advocate of reason. He was highly involved with the question of the truth. What is true and what is wrong? What is the way of a true life? Socrates had no ready answers. He left it to each of his students themselves, to find their own way of truth. Three things he gave them along the way: 1. Keep interested in the truth. 2. Make sure that your soul is as good as possible. 3. To get a good soul, maintain the four virtues of prudence, temperance, courage and justice (charity).(Plato, Apologie, Stuttgart 1982, p. 36)

Aristippus of Cyrene
Perhaps the first philosopher who has developed a complete philosophy of happiness was Aristippus. He was a student of Socrates, but adopted a very different philosophical outlook, teaching that the goal of life was to seek external pleasure. Aristippus lived luxuriously. He is considered the founder of hedonism.

Antisthenes
Antisthenes (c. 445 BCE – c. 365 BCE) was also a student of Socrates. He adopted and developed the ethical side of Socrates' teachings, advocating an ascetic life lived in accordance with virtue. Later writers regarded him as the founder of Cynic philosophy. His most important disciple was Diogenes, who lived after a legend in a barrel. The way of happiness of Antisthenes is similar to the Enlightenment philosophy of Buddhism, Indian Yoga and Chinese Taoism. Through a life of peace, simplicity, naturalness, modesty and virtue (mental work) dissolve the inner tensions. Inner happiness and enlightenment appear. We find Antisthenes praising the pleasures which spring "from out of one's soul."(Xenophon, Symposium, iv. 41.)

Plato
Plato lived from 428/427 BC to 348/347 BC in Athens. He was a student of Socrates, writer of philosophical dialogues, and founder of the Academy in Athens, the first institution of higher learning in the Western world. According to Plato the human soul consists of three parts: The reason, the will and the desire. A man is happy when all three parts of the soul are in balance. Plato has thought about how to build a good society. He proposed to transfer the leadership of a society to the wise. One could say that science (the philosophy of happiness) should be the center of happy society.

A student of Plato was Aristotle. According to Aristotle, happy is he who develops his virtues and abilities. A man can be called perfectly happy if he is sufficiently equipped with external goods and spends his life according to virtue.(Nicomachean Ethics Book I.)

Epicurus
For Epicurus, the purpose of philosophy was to attain a happy, tranquil life, characterized by peace, freedom from fear, the absence of pain, and by living a self-sufficient life surrounded by friends. A life after Epicurus (341-270 BC) is happy when you live everything in the right degree. Everyone should know his point of enough. "Whom is enough too little, nothing is enough."(Johannes Mewaldt: Epikur. Philosophie der Freude. Stuttgart 1973, page 71.)

Epicureans often confounded with the hedonists. Both are completely different philosophical paths. An Epicurean embodies a moderate path of asceticism and a hedonist a path of extreme external pleasure. Epicureanism is wisdom and hedonism is unwisdom. Epicureanism leads to enlightenment (inner happiness) and hedonism to unenlightenment (inner tensions, addictions).

Epicurus taught positive thinking. A life will be happy when we constantly train positive thinking. Epicurus called it "philosophize." A person should philosophize every day. One should think about the meaning of life and reflect again and again to his positive goals. One should avoid it, to worry too much.

The inner happiness comes from inner peace. When a person calms down, inner happiness appears. Epicurus recommended it to live in inner peace, "Then you live like a God (Buddha) among your unwise fellow men,"(Johannes Mewaldt, ibid, page 48.) which an Epicurean also helps on the way: "The friendship dances around the globe, all of us announcing that we shall awaken to bliss."(Johannes Mewaldt, ibid, page 70.)

Enlightenment
Enlightenment is the central point to understand the philosophy of happiness. According to Swami Sivananda, there are three main paths to enlightenment, the path of peace (live in rest), the path of love (do good) and the path of practicing (yoga, meditation). The best way is to practice all three ways.

Everyone should find the spiritual exercises that suit him well. In order to resolve our internal tensions and to awaken our inner happiness, we need exercises for the body (yoga, walking, meditation) and exercises for the mind (self-reflection, thoughts control, doing good). Basically you get to the enlightenment (to a life in the light / in God) through the personally right combination from meditation and mental work.

The basic principle of Epicurus was, "A philosopher lives like a God (Buddha, Enlightened) among the people." The most important enlightenment technique for Epicurus was next to a life in stillness the daily positive thinking. Epicurus called it "philosophize." A life will be happy when we constantly think positive and live with enough rest. A person should philosophize every day. He should think about the meaning of life, reflect on his positive goals and consistently go his path of truth and wisdom.

Undemandingness
Epicurus taught to be frugal in external things and to focus on the inner happiness. His goal was to overcome all fears and to live as a God (Buddha) among men. He recommended the path of small pleasure. He was famous for the phrase, "Send me a small piece of cheese, then I can really enjoy."

Nils was from the beginning of his life in search of the great good fortune. He was looking for happiness in relationships, career and beautiful journeys. When he was thirty years old, there was a good opportunity to deepen his knowledge. He stood at the end of his legal studies. The written work of the exam had been completed. Now he had to wait three months until the inspectors had evaluated the work. These three months he was free. He wondered what he should do with this many leisure. Nils remembered his previously unsuccessful search for lasting happiness. He came up with the idea to read all books about happiness systematically. Maybe the happiness books could help him in his personal search?

He scoured in the libraries for scientific happiness literature. He looked through the whole literature on the word "happiness". He bought all the books that seemed to be helpful in this quest. Overall, he bought some twenty books. Then he retired to his study and read them all consecutively. At first he was confused by the diversity of views about happiness. But after about two months ago, he saw things more clearly. He recognized that some authors were on the wrong track. Most authors were groping in the dark in their search for happiness. They had actually understood anything. Nils recognized this by comparing their results with his own experiences of happiness. Some authors, however, had an eye for the true way. They watched the people who are happy in their lifes. And found that these people are essentially characterized by two special qualities: their positivity and their modesty. The most happiest men first thought positive and secondly, had a modest nature.

That positive thinking contributes to inner happiness is understandable. If you think positive, positive feelings are generated in your psyche. This is also confirmed by the happiness research. But what is going on with the property "modesty"? This property is at the present time very forgotten. We live in a first-person company. Modesty is considered to be unwise. Immodesty is the great idea of life. The more you want, the more you're gonna get. The problem of today's doctrine of immodesty is that the goal of life is located on the wrong place. The goal of life is seen outside. Today's consumer prophets suggest that maximum external pleasure automatically leads to maximum inner happiness. This is a false doctrine that is clearly contradicted by reality.

In particular, this approach does not work long term. May the outward pleasure be as great as it wants. After some time the mind becomes accustomed and will hardly notice it as something special. Even large external wealth does not make you happy on a deep level in the long run. You even get used to an attractive partner and good regular sex. You seek for decades after Prince Charming or work for material wealth. And then all additional happiness disappears a short time after you have reached your goals. Today's happiness research has found out that a sufficient material basis of life is important for the inner happiness. But then inner happiness is not growing more by external goods, but by psychological factors.

The pursuit of external wealth, professional success and the perfect soul mate turns out to be always after a while as a mistake. Some time after happy external events every man turns back to his normal mental happiness level. Who in his life wants to be happy at a deep level must focus on his inner happiness. He must reduce his inner tensions and maintain qualities such as inner peace, love and positivity. The real goal of life is the greatest possible inner happiness. We have to be immodest at inner happiness. We must strive for enlightenment. But we must be modest in outer pleasure. Most people confuse these two things. In Yoga we consider the more inner peace as the basis of happiness as positive thinking. Who is relaxed internally automatically thinks positive. This is the main direction of Yoga. Who wants to force his inner happiness by positive thinking easily gets inner tensions. He is only superficially positive. Anyone who is enlightened, is positive deeply in his soul.

Objectivism, the philosophy of Ayn Rand
Rand believed human beings, and the rest of the universe, all follow the The Law of Identity. Happiness is a person's emotional state when they are living in a manner aligned to their own identity, and the identity of their environment. For example, certain foods and drinks effect a person positively, and others negatively. If a person purposefully ingests the negative things, that will act against his life, and he will feel unhappy about it since he did it knowingly.

Now, perhaps another person likes to be sociable, have many friends, and contribute to the well-being of her community. Her psychological nature, i.e. identity, has this social requirement. If that requirement is met, it will contribute to her happiness.

Sometimes bad things happen in life, and people suffer. There are diseases with no cure. Storms wreck villages, and loved ones are lost. Such occurrences are no ones fault. As long as an individuals' choices are aligned with the Law of Identity, they are honest and true. The suffering will only go so deep. Underneath the grief, the individual will still feel fit to live. If, on the other hand, the person was attempting to defy the Law of Identity, eventually it will catch him, and he will know that it was his own doing, and that he is not fit. That's when the suffering reaches the core of a person.

An outstanding example of how people can be happy in a harsh environment is the true story of "Shackleton's Incredible Voyage".

Consciousness is part of the universe. Just like everything else, it also follows the Law of Identity. Science, and Physics rest on this law, along with a belief in Reason as the means to knowledge. These disciplines continue to explain more and more about how phenomena work. See the The Universal Law of Happiness for additional discussion on the physics of happiness.

What is Your Way?
The center of philosophy is the question of the meaning of life. Who goes through life without this issue, lives unconscious. A person is one only by thinking philosophical. Everyone should be a philosopher. Blessed is the one, who finds his purpose of life through his philosophical thinking. He can live wisely and strategically. He will receive a full life. Live happily. Live with the happiness research and the philosophy of happiness.

Basically, the way of the outer pleasure and the way of the inner happiness are opposites. Follow the right principles of life. The way of the outer pleasure creates tensions (a big ego, wishes, inner stress) and you become inwardly unhappy. The path of the outer pleasure causes at the long-term a growth into the inner unhappiness. We are never internally really satisfied, and thus we seek to go more extreme the way of the external pleasure. Until we recognize its failure.

On the path of inner happiness, all inner tensions dissolve. We systematically resolve the tensions from the body and the mind. If the spirit becomes free of his fears (attachment to suffering situations) and desires (attachment to external pleasures), inner peace appears. We get into a cosmic consciousness. Lasting inner happiness evolves. We are content with ourselfes and our lifes. We have released all external things at a deeper level and thus realize our true self.

The people are different and have different ways of inner happiness. For some people it is good to live as an extreme ascetic (Diogenes). For some people it is right to try the path of the outer pleasure enough, until they are ready for the path of inner happiness. For most people, a mean way with some outer joy is the best spiritual way.

1. Think about life. What are your goals? What do you want to reach in your life? "My goals are ... "

2. Think about inner and outer happiness. The happiness research found out, that happiness comes to 90% from inside and only to 10 % from outer things like a great partner, much money or a big career (Sonja Lyubomirsky). As a philosopher invest your energy in a healthy life, in meditation, in positive thinking and in working for a happy world.

3. Think about enlightenment. Enlightenment is the biggest happiness. Find your personal way of spirituality. Center your life in Logic (in a higher reason like truth, love, inner happiness or enlightenment). What are the spiritual exercise you want to do daily (yoga, meditation, reading, positiv thinking, walking, praying, doing good)?

4. Make a good plan of your life. In yoga there are four phases of a full life, learning in the childhood, work as adult, relationship or family and selfrealization.

5. Live as a philosopher of happiness. You can go the fast way to enlightenment like Diogenes (live like a yogi), the middle way to enlightenment like Epicurus (live inner and outer happiness at the same time) or the way of unenlightenment like Aristippus (outer pleasure and inner suffering). Test the best!!

title: Happiness/Philosophy of Happiness


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User talk:Nobody60

 Hello Nobody60, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --mikeu talk 16:15, 24 March 2011 (UTC)

Thank you.--Nobody60 16:18, 24 March 2011 (UTC)

Image uploads
I've noticed a few of the images you've uploaded are also on Wikimedia Commons. If you do find an image there which you want to use you can just refer to the filename in your page here and it will appear, you don't need to upload it specifically to Wikiversity. Regards. Adambro 13:24, 3 January 2012 (UTC)
 * Thank you. --Nobody60 13:30, 3 January 2012 (UTC)

Tarot
I've revised this tarot article as it was too narrow in scope. Readers are mislead into believing the card reading is the only use of tarot cards. For one thing it makes no mention of the fact that tarot cards were really made for card games still played today in some countries. Tarot articles for something like Wikiversity should take a more global perspective and not simply give interpretations of the cards. The older version has been preserved with a more appropriate title. A Psychological Interpretation of the TarotSmiloid (talk) 20:50, 8 March 2012 (UTC)
 * I've decided to merge the two articlesSmiloid (talk) 21:00, 8 March 2012 (UTC)

Okay.--Nobody60 (talk) 09:26, 9 March 2012 (UTC)

Your essays or articles
You have created a series of pages on Wikiversity that appear to be personal essays, rather than general educational resources. On the face of it, I'd think of moving these to your user space, but some of them could be subpages of resources on religion or theology or yoga, i.e., participant essays.

Wikiversity does have a neutrality policy, but we generally approach neutrality inclusively, rather than excluding material that expresses a personal point of view. A page sitting in mainspace that is clearly written from a personal point of view, rather than being written from a neutral point of view, invites opposition from users accustomed to Wikipedia practice, and it can be contrary to the Wikiversity mission if not handled with care,

What do you think? --Abd (discuss • contribs) 21:45, 18 October 2013 (UTC)

Image Tagging Image:Mariatu.jpg
Thanks for uploading Image:Mariatu.jpg. I notice the image page currently doesn't specify who created the content, so the copyright status is unclear. If you have not created this media yourself then you need to argue that we have the right to use the media on Wikimedia Commons (see copyright tagging below). If you have not created the media yourself then you should also specify where you found it, i.e., in most cases link to the website where you got it, and the terms of use for content from that page. If the content is a derivative of a copyrighted work, you need to supply the names and a licence of the original authors as well.

If the media also doesn't have a copyright tag, then you must also add one. If you created/took the picture, audio, or video then you can use to release it under the multilicense GFDL plus Creative Commons Attribution-ShareAlike All-version license or PD-self to release it into the public domain. See commons:Commons:Copyright tags for the full list of copyright tags that you can use.

Note that any unsourced and untagged images will be deleted one week after they have been uploaded, as described on criteria for speedy deletion. If you have uploaded other media, please check that you have specified their source and copyright tagged them, too. You can find all your uploads using the Gallery tool. Thank you. TeleComNasSprVen (discuss • contribs) 06:36, 15 March 2014 (UTC)
 * I don't need the image. So you can delete it. --Nobody60 (discuss • contribs) 11:47, 15 March 2014 (UTC)
 * See my comment below. Because you did not state that you are the creator, I have not edited the page as I did with that other file. You may either request deletion, allow usage by claiming authorship/ownership, or do nothing and it will likely be deleted. --Abd (discuss • contribs) 14:35, 15 March 2014 (UTC)

File Tagging File:Ulla engel.JPG
Thanks for uploading File:Ulla engel.JPG. I notice the image page currently doesn't specify who created the content, so the copyright status is unclear. If you have not created this media yourself then you need to argue that we have the right to use the media on Wikimedia Commons (see copyright tagging below). If you have not created the media yourself then you should also specify where you found it, i.e., in most cases link to the website where you got it, and the terms of use for content from that page. If the content is a derivative of a copyrighted work, you need to supply the names and a licence of the original authors as well.

If the media also doesn't have a copyright tag, then you must also add one. If you created/took the picture, audio, or video then you can use to release it under the multilicense GFDL plus Creative Commons Attribution-ShareAlike All-version license or PD-self to release it into the public domain. See commons:Commons:Copyright tags for the full list of copyright tags that you can use.

Note that any unsourced and untagged images will be deleted one week after they have been uploaded, as described on criteria for speedy deletion. If you have uploaded other media, please check that you have specified their source and copyright tagged them, too. You can find all your uploads using the Gallery tool. Thank you. TeleComNasSprVen (discuss • contribs) 08:49, 15 March 2014 (UTC)
 * I don't need the image. So you can delete it. But I have the copyright. --Nobody60 (discuss • contribs) 11:48, 15 March 2014 (UTC)
 * It is best if you make the choice yourself, Nobody60. The information above referred to Commons, but this is a local file. You may request the image be deleted by placing at the top of the information page. However, in case you want the keep the image for future use, and/or allow others to use it, I have added information from your comment above, assuming that "I have the copyright" means that the image is your own work. If not, please let me know or revert that edit. Thanks. --Abd (discuss • contribs) 14:30, 15 March 2014 (UTC)
 * In copyright analysis, asserting that one has the copyright is different from asserting that it is the product of one's own work. The source is needed to verify that the uploader has obtained the copyright from the creator necessary for licensing the media file under one of Wikiversity's free licenses, or otherwise. TeleComNasSprVen (discuss • contribs) 20:10, 15 March 2014 (UTC)
 * TCNSV may be technically correct, for a Commons page. He might not be correct for here. If you want to nail this down, Nobody60, you would edit the File page and remove my comment, which points here, and replace it with a simple "own work" or whatever else is the actual case. Or you would just write here "I made this" or something like that. --Abd (discuss • contribs) 20:30, 15 March 2014 (UTC)

Spiritual Almanac
I have moved your personal essay or blog ("almanac") to User:Nobody60/Spiritual Almanac. That created a redirect which may eventually be deleted, so make sure that any off-wiki links you create point to the user subpage. Thanks for contributing to Wikiversity. --Abd (discuss • contribs) 20:51, 26 July 2014 (UTC)

Files Missing Source Information
The following files are missing source information. Source information is required for files to remain at Wikiversity. Please edit these files and add the missing information. Let me know if you have any questions.

 * File:Altes Futhark.png
 * File:Angel of Forgiveness.jpg
 * File:Baum im Sturm.JPG
 * File:Bede Griffiths.jpg
 * File:Blumenhaus.JPG
 * File:Buddha silent in the flowing time.png
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 * File:Justice.jpg
 * File:Little Nils.jpeg
 * File:Magician.jpg
 * File:Mahaprabhuji1.jpeg
 * File:Nils 2012.JPG
 * File:Nils Yogateacher.jpg
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 * File:Nils reitet auf einem Besen.jpg
 * File:Nils und eine Freundin.jpeg
 * File:Nils überschreitet eine Brücke.jpeg
 * File:Nils2012.JPG
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 * File:Resurrection.jpg
 * File:Self Mastery.jpg
 * File:Shantimayi mit Meister.jpg
 * File:Shri Lakshmi herself.jpg
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 * File:Tarot 13 Death.jpg
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 * File:Temperance.jpg
 * File:The Chariot.jpg
 * File:The Star.jpg
 * File:The Sun (Tarot).jpg
 * File:The Tower.jpg
 * File:The cancer.jpg
 * File:Tridevi.png
 * File:Ulla2.jpg
 * File:Wands01.jpg
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 * File:Wheel of Fortune.jpg
 * File:Yogalehrer.jpg
 * File:Yogi Nils mit Tannenbaum.JPG
 * File:Yogi Prahlad Jani.jpg
 * File:YogiKatze.JPG
 * File:Yogin Giri Bala.jpg

Dave Braunschweig (discuss • contribs) 03:59, 14 April 2016 (UTC)

The Tarot cards are free of copyright and are from the Internet. All other pictures are mine. You can paste the information. Or I'll do it when I have time and inclination.--Nobody60 (discuss • contribs) 14:32, 14 April 2016 (UTC)

 * Unfortunately, according to Wikipedia:File:Cups01.jpg, the Tarot cards are not yet free worldwide, and when they would be free depends on the source, so a source is required in order for them to remain at Wikiversity. The pictures that appeared to be yours were credited accordingly. The pictures that were identified as copyrighted were deleted. The pictures that have multiple potential sources have been tagged as needing license information, and will be deleted if information is not provided. Let me know if you have any questions. -- Dave Braunschweig (discuss • contribs) 16:17, 14 April 2016 (UTC)

Files Missing Information
Thanks for uploading files to Wikiversity. All files must have source and license information to stay at Wikiversity. The following files are missing and/or License tags, and will be deleted if the missing information is not added. See Uploading files for more information.

 * File:Bede Griffiths.jpg
 * File:Blumenhaus.JPG
 * File:Cups01.jpg
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 * File:God (Light, Guardian Angel).jpg
 * File:Hanged Man.jpg
 * File:Hermit.jpg
 * File:Herrscherin.jpg
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 * File:Justice.jpg
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 * File:Wands06.jpg
 * File:Wands07.jpg
 * File:Wands08.jpg
 * File:Wands09.jpg
 * File:Wands10.jpg
 * File:Wands11.jpg
 * File:Wands12.jpg
 * File:Wands13.jpg
 * File:Wands14.jpg
 * File:Wheel of Fortune.jpg
 * File:Yogi Prahlad Jani.jpg
 * File:Yogin Giri Bala.jpg

MaintenanceBot (discuss • contribs) 15:30, 16 April 2016 (UTC)

title: User talk:Nobody60