text:
User:Eml5526.s11.team01/Hwk2

=Problem 1=

Given
An elastic bar with a cross section A(x)

Find
Derive heat problem equation from Lecture Slide Mtg.6
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$$ \frac{\partial}{\partial x}\left(A \left(x\right) k\left(x\right) \frac{\partial u}{\partial x}\right) + f \left(x,t \right) = A \left(x \right) \rho c \frac{\partial u}{\partial t} $$

 * $$\displaystyle (Eq. 1.1) $$
 * }
 * }

Solution
Fig. 1 Shows a bar of cross section A(x), with conductivity k(x), density ρ, specific heat c, with an internal heat source (or sink) f(x,t). In the bar the heat is transferred only by conduction in x direction. The temperature has been called u(x,t)

The partial differential equation for this problem can be obtained doing an energy balance in an infinitesimal element of the bar. Fig. 2 shows an infinitesimal element of length dx; heat entering the body on the left hand side is $$q\left(x \right)$$ and heat going out the body on the right hand side is $$q \left(x + dx \right)$$

Doing the energy balance we obtain:
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$$ q \left(x \right) A \left(x \right) - q \left(x + dx \right) A \left(x+dx \right) + fdx = \frac{dE}{dt}$$
 * $$\displaystyle (Eq. 1.2) $$
 * }
 * }

Where $$ \frac{dE}{dt}$$ is the change of energy with respect to time which can be calculated in terms of the temperature as:
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$$ \frac{dE}{dt} = A d x \rho c \frac{\partial u}{\partial t} $$
 * $$\displaystyle (Eq. 1.3) $$
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The heat flow and the cross area at the right hand side of the element can be approximated by:
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$$ q \left(x + dx \right) = q \left(x \right) + \frac{\partial q}{\partial x} dx $$
 * $$\displaystyle (Eq. 1.4) $$
 * }
 * }

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$$ A \left(x + dx \right) = A \left(x \right) + \frac{\partial A}{\partial x} dx $$
 * $$\displaystyle (Eq. 1.5) $$
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 * }

Replacing on Eq.1.2 we obtain:
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$$ qA - \left(q + \frac{\partial q}{\partial x}dx \right)\left(A + \frac{\partial A}{\partial x} dx \right) + fdx = A d x \rho c \frac{\partial u}{\partial x} $$
 * $$\displaystyle (Eq. 1.6) $$
 * }
 * }

Doing the product we get:
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$$ qA - qA - q \frac{\partial A}{\partial x} dx - A \frac{\partial q}{\partial x} dx - \frac{\partial A}{\partial x} dx \frac{\partial q}{\partial x} dx + fdx = A d x \rho c \frac{\partial u}{\partial x} $$
 * $$\displaystyle (Eq. 1.7) $$
 * }
 * }

Neglecting the second order term we have:
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$$ -q \frac{\partial A}{\partial x} dx - A \frac{\partial q}{\partial x} dx + fdx =Adx \rho c \frac{\partial u}{\partial x}$$
 * $$\displaystyle (Eq. 1.8) $$
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 * }

The first two terms in the latter equation are the derivative of a product; therefore:
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$$ - \frac{\partial}{\partial x} \left(Aq \right) dx + fdx = Adx \rho c \frac{\partial u}{\partial x} $$
 * $$\displaystyle (Eq. 1.9) $$
 * }
 * }

Using the Fourier’s law, the heat flow can be expressed as function of temperature. For an unidirectional case we have:
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$$ q\left (x \right) = -k \left(x \right) \frac{\partial u}{\partial x} $$
 * $$\displaystyle (Eq. 1.10) $$
 * }
 * }

Where k(x) is the conductivity of the material. Replacing equation Eq.1.9 in Eq.1.10 we have:
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$$ \frac{\partial}{\partial x} \left( Ak \frac{\partial u}{\partial x} \right) + f \left(x,t \right) = A \rho c \frac{\partial u}{\partial t} $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 * $$\displaystyle ( Eq. 1.11) $$
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=Problem 2=

Given
$$\left [B _{jk} \right ]=\begin{bmatrix} 1 &1 &1 \\ 2 &-1 &3 \\ 3 &2  &6 \end{bmatrix}$$
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$$\overset{v}{\rightarrow}=5\overset{a_{1}}{\rightarrow}-7\overset{a_{2}}{\rightarrow}-4\overset{a_{3}}{\rightarrow}$$

 * $$\displaystyle (Eq. 2.1) $$
 * }
 * }

Find
2.1 $$ Det\left [B _{jk} \right ]$$.

2.2 Find $$\displaystyle \Gamma (b_{1},b_{3},b_{3})=K$$.

2.3 Find $$\overset{F}{\rightarrow}=\left \{ F_{i} \right \}=\left \{ b_{i}.v \right \}$$.

2.4 Solve $$\displaystyle (Eq. 5) $$ on 7-2 for $$\overset{d}{\rightarrow}=\left \{v _{j} \right \}$$

2.5 Use equation 1 on 7-4 to find $$\vec{\overset{k}{\rightarrow}}\overset{d}{\rightarrow}=\vec{\overset{F}{\rightarrow}}$$ What is $$\vec{K}$$ & $$\vec{F}$$?

2.6 Solve for $$\overset{d}{\rightarrow}_{i}$$ compare to $$\overset{d}{\rightarrow}$$ in 4).

2.7 Observe the symmetric properties of $$\overset{K}{\rightarrow}$$ & $$\overset{\vec{K}}{\rightarrow}$$. Further Discuss the Pros & Cons of Bubnov-Galerkin & Petrov-Galerkin Methods.

2.1
$$ Det\left [B _{jk} \right ]$$=
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$$\begin{vmatrix} 1 &1 &1 \\ 2 &-1 &3 \\ 3 &2  &6 \end{vmatrix}=-8.$$.
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 * $$$$
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2.2
Find  $$\Gamma (b_{1},b_{3},b_{3})=\overset{K}{\rightarrow}$$.

According to the $$\displaystyle Eq.4 $$ on 7-1, $$b_{1},b_{3},b_{3}$$ given before we have,

$$\overset{b_{j}}{\rightarrow}.d_{j}=\overset{V}{\rightarrow}$$

Successively performing multiplication by $$b_{1},b_{3},b_{3}$$  we get,
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$$d_{1}\overset{b_{1}}{\rightarrow}+d_{2}\overset{b_{2}}{\rightarrow}+d_{3}\overset{b_{3}}{\rightarrow}=\overset{V}{\rightarrow}$$

 * $$\displaystyle (Eq. 2.2) $$
 * }
 * }

Substituting the given values we get,

$$d_{1}\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}+d_{2}\begin{bmatrix} 2\\ -1\\ 3 \end{bmatrix}+d_{3}\begin{bmatrix} 3\\ 2\\ 6 \end{bmatrix}=\begin{bmatrix} 5\\ -7\\ -4 \end{bmatrix}$$

This equation can also be written in the following format,

$$\begin{bmatrix} 1 &2 &3 \\ 1 &-1  &2 \\ 1 & 3 & 6 \end{bmatrix}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{pmatrix} 5\\ -7\\ 4 \end{pmatrix}$$

$$\displaystyle Eq.3 $$ on 7-2 is obtained by multiplying $$\displaystyle Eq.2.2  $$ by each $$\overset{b}{\rightarrow}$$, so we can write,

$$\overset{b_{1}}{\rightarrow}.\overset{b_{1}}{\rightarrow}d_{1}+\overset{b_{1}}{\rightarrow}.\overset{b_{2}}{\rightarrow}d_{2}+\overset{b_{1}}{\rightarrow}.\overset{b_{3}}{\rightarrow}d_{3}=\overset{b_{1}}{\rightarrow}.\overset{V}{\rightarrow}==F_{1}$$

$$\overset{b_{2}}{\rightarrow}.\overset{b_{1}}{\rightarrow}d_{1}+\overset{b_{2}}{\rightarrow}.\overset{b_{2}}{\rightarrow}d_{2}+\overset{b_{2}}{\rightarrow}.\overset{b_{3}}{\rightarrow}d_{3}=\overset{b_{2}}{\rightarrow}.\overset{V}{\rightarrow}==F_{2}$$

$$\overset{b_{3}}{\rightarrow}.\overset{b_{1}}{\rightarrow}d_{1}+\overset{b_{3}}{\rightarrow}.\overset{b_{2}}{\rightarrow}d_{2}+\overset{b_{3}}{\rightarrow}.\overset{b_{3}}{\rightarrow}d_{3}=\overset{b_{2}}{\rightarrow}.\overset{V}{\rightarrow}==F_{3}$$

These 3 equations can be further presented as below in the matrix form,

$$\begin{bmatrix} \overset{b_{1}}{\rightarrow}.\overset{b_{1}}{\rightarrow} & \overset{b_{1}}{\rightarrow}.\overset{b_{2}}{\rightarrow} & \overset{b_{1}}{\rightarrow}.\overset{b_{3}}{\rightarrow}\\ \overset{b_{2}}{\rightarrow}.\overset{b_{1}}{\rightarrow} & \overset{b_{2}}{\rightarrow}.\overset{b_{2}}{\rightarrow} & \overset{b_{2}}{\rightarrow}.\overset{b_{3}}{\rightarrow}\\ \overset{b_{3}}{\rightarrow}.\overset{b_{1}}{\rightarrow} & \overset{b_{3}}{\rightarrow}.\overset{b_{2}}{\rightarrow} & \overset{b_{3}}{\rightarrow}.\overset{b_{3}}{\rightarrow} \end{bmatrix}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{bmatrix} b_{1}\\ b_{2}\\ b_{3} \end{bmatrix}\begin{pmatrix} V_{1}\\ V_{2}\\ V_{3} \end{pmatrix}$$

Which also can be written as

$$\begin{bmatrix} \overset{b_{1}}{\rightarrow}\\ \overset{b_{2}}{\rightarrow}\\ \overset{b_{3}}{\rightarrow} \end{bmatrix} \begin{bmatrix} \overset{b_{1}}{\rightarrow}\\ \overset{b_{2}}{\rightarrow}\\ \overset{b_{3}}{\rightarrow} \end{bmatrix}^{T}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{bmatrix} \overset{b_{1}}{\rightarrow}\\ \overset{b_{2}}{\rightarrow}\\ \overset{b_{3}}{\rightarrow} \end{bmatrix}\begin{pmatrix} V_{1}\\ V_{2}\\ V_{3} \end{pmatrix}=\overset{F}{\rightarrow}$$

Substituting the numerical values we get,

$$\begin{bmatrix} 1 &1 &1 \\ 2 &-1 &3 \\ 3 &2  &6 \end{bmatrix}\begin{bmatrix} 1 &2 &3 \\ 1 &-1 &2\\ 1 &3  &6 \end{bmatrix}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{bmatrix} 1 &1 &1 \\ 2 &-1 &3 \\ 3 &2  &6 \end{bmatrix}\begin{bmatrix} 5\\ -7\\ -4 \end{bmatrix}$$

Performing the product, we get:
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$$\begin{bmatrix} 3 &4 &11\\ 4 &14  &22\\ 11 &22  &49 \end{bmatrix}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{bmatrix} 1 &1 &1 \\ 2 &-1 &3 \\ 3 &2  &6 \end{bmatrix}\begin{bmatrix} 5\\ -7\\ -4 \end{bmatrix}=\overset{F}{\rightarrow}$$

 * $$\displaystyle (Eq. 2.3 ) $$
 * }
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$$\displaystyle Eq. 2.3 $$ is equivalent to $$\displaystyle Kd=F$$ where
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 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

$$K=\begin{bmatrix} 3 &4 &11 \\ 4 &14 &22 \\ 11&22 &49 \end{bmatrix}=Stiffness Matrix$$

 * $$$$
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 * }

Furthermore,
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$$Det[K]=\begin{vmatrix} 3 &4 &11\\ 4 &14 &22 \\ 11 &22 &49 \end{vmatrix}= 64$$
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 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 *
$$$$
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2.3
$$\overset{F}{\rightarrow}=\left \{ F_{i} \right \}=\left \{ b_{i}.v \right \}$$=

From $$\displaystyle Eq. 2.3 $$
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$$\overset{F}{\rightarrow}=\begin{bmatrix} 1 &1 &1 \\ 2 &-1 &3 \\ 3 &2  &6 \end{bmatrix}\begin{bmatrix} 5\\ -7\\ -4 \end{bmatrix}=\begin{pmatrix} -6\\ 5\\ -23 \end{pmatrix}$$

 *
$$$$
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 * }

2.4
Solve $$\displaystyle Eq. 5 $$ on 7-2 for $$\overset{d}{\rightarrow}=\left \{v _{j} \right \}$$

Rearranging $$\displaystyle Eq. 2.3 $$  We get

$$\displaystyle d=K^{-1}F$$
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$$d=\begin{bmatrix} 3 &4 &11 \\ 4 &14  &22 \\ 11 &22  &49 \end{bmatrix}^{-1}\begin{pmatrix} -6\\ 5\\ -23 \end{pmatrix}=\begin{pmatrix} 8.3750\\ 5.6250\\ -4.875 \end{pmatrix}$$
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 *
$$$$
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2.5
Use equation 1 on 7-4 to find $$\vec{\overset{k}{\rightarrow}}\overset{d}{\rightarrow}=\vec{\overset{F}{\rightarrow}}$$. Further, What is $$\vec{K}$$ & $$\vec{F}$$?

In this case, $$\displaystyle w_{i}=a_{i}$$, where $$\displaystyle a_{i}$$ is the orthonormal basis.

Starting with Equation 1 on 7-2, we get:

$$d_{1}\overset{b_{1}}{\rightarrow}+d_{2}\overset{b_{2}}{\rightarrow}+d_{3}\overset{b_{3}}{\rightarrow}=\overset{V}{\rightarrow}$$

If we multiply successively by $$\displaystyle a_{i}$$, we get the following equations,

$$\overset{a_{1}}{\rightarrow}.\overset{b_{1}}{\rightarrow}d_{1}+\overset{a_{1}}{\rightarrow}.\overset{b_{2}}{\rightarrow}d_{2}+\overset{a_{1}}{\rightarrow}.\overset{b_{3}}{\rightarrow}d_{3}=\overset{a_{1}}{\rightarrow}.\overset{V}{\rightarrow}=F_{1}$$

$$\overset{a_{2}}{\rightarrow}.\overset{b_{1}}{\rightarrow}d_{1}+\overset{a_{2}}{\rightarrow}.\overset{b_{2}}{\rightarrow}d_{2}+\overset{a_{2}}{\rightarrow}.\overset{b_{3}}{\rightarrow}d_{3}=\overset{a_{2}}{\rightarrow}.\overset{V}{\rightarrow}=F_{2}$$
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$$\overset{a_{3}}{\rightarrow}.\overset{b_{1}}{\rightarrow}d_{1}+\overset{a_{3}}{\rightarrow}.\overset{b_{2}}{\rightarrow}d_{2}+\overset{a_{3}}{\rightarrow}.\overset{b_{3}}{\rightarrow}d_{3}=\overset{a_{3}}{\rightarrow}.\overset{V}{\rightarrow}=F_{3}$$

 *
$$\displaystyle (Eq. 2.4) $$
 * }
 * }

Which can be further written as,

$$\begin{bmatrix} \overset{a_{1}}{\rightarrow}\\ \overset{a_{2}}{\rightarrow}\\ \overset{a_{3}}{\rightarrow} \end{bmatrix}\begin{bmatrix} \overset{b_{1}}{\rightarrow}\\ \overset{b_{2}}{\rightarrow}\\ \overset{b_{3}}{\rightarrow} \end{bmatrix}^{T}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{bmatrix} \overset{a_{1}}{\rightarrow}\\ \overset{a_{2}}{\rightarrow}\\ \overset{a_{3}}{\rightarrow} \end{bmatrix}\begin{pmatrix} V_{1}\\ V_{2}\\ V_{3} \end{pmatrix}=\overset{\vec{F}}{\rightarrow}$$

As $$ a_{i}$$ is orthonormal basis, we can write the following,

$$a_{1}=\begin{bmatrix} 1 & 0 &0 \end{bmatrix};a_{2}=\begin{bmatrix} 0&1&0 \end{bmatrix};a_{3}=\begin{bmatrix} 0& 0 &1 \end{bmatrix}$$

Substituting the respective values we have,

$$\begin{bmatrix} 1 &0 &0 \\ 0 &1  &0 \\ 0 &0  &1 \end{bmatrix}\begin{bmatrix} \overset{b_{1}}{\rightarrow}\\ \overset{b_{2}}{\rightarrow}\\ \overset{b_{3}}{\rightarrow} \end{bmatrix}^{T}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0& 1 \end{bmatrix}\begin{pmatrix} V_{1}\\ V_{2}\\ V_{3} \end{pmatrix}=\overset{\vec{F}}{\rightarrow}$$

Simplifying we obtain that,

$$\begin{bmatrix} \overset{b_{1}}{\rightarrow}\\ \overset{b_{2}}{\rightarrow}\\ \overset{b_{3}}{\rightarrow} \end{bmatrix}^{T}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{pmatrix} V_{1}\\ V_{2}\\ V_{3} \end{pmatrix}=\overset{\vec{F}}{\rightarrow}$$

Replacing the numerical values we get,
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$$\begin{bmatrix} 1 & 2 & 3\\ 1 &-1 &2 \\ 1 &3  & 6 \end{bmatrix}\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{pmatrix} 5\\ -7\\ -4 \end{pmatrix}$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 *
$$\displaystyle ( Eq. 2.5 ) $$
 * style= |
 * }

This equation is nothing but the system

$$\overset{\vec{K}}{\rightarrow}\overset{d}{\rightarrow}=\overset{\vec{F}}{\rightarrow}$$

From where we can conclude that $$\overset{\vec{K}}{\rightarrow}$$ is a matrix whose columns are the vectors $$\overset{b_{j}}{\rightarrow}$$ and $$\overset{\vec{F}}{\rightarrow}= V$$.

2.6
Solve for $$\overset{d}{\rightarrow}_{i}$$ compare to $$\overset{d}{\rightarrow}$$ in problem 2.4
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$$\begin{pmatrix} d_{1}\\ d_{2}\\ d_{3} \end{pmatrix}=\begin{bmatrix} 1 & 2 & 3\\ 1 &-1 &2 \\ 1 &3  & 6 \end{bmatrix}^{-1}\begin{pmatrix} 5\\ -7\\ -4 \end{pmatrix}=\begin{pmatrix} 8.3750\\ 5.6250\\ -4.875 \end{pmatrix}$$
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 *
$$$$
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We can see that this result is identical to the result we obtained in problem 2.4

2.7
Observe the symmetric properties of $$\overset{K}{\rightarrow}$$ & $$\overset{\vec{K}}{\rightarrow}$$. Further Discuss the Pros & Cons of Bubnov-Galerkin & Petrov-Galerkin Methods.

We can explicitely see that $$\overset{K}{\rightarrow}$$  is always symmetric because it is obtained from the product of $$\overset{b_{j}}{\rightarrow}.\overset{b_{i}}{\rightarrow}$$; where as $$\overset{\vec{K}}{\rightarrow}$$ is not always symmetric. It will be symmetric only when the vector b is a Orthonormal basis.

The pros of Bubnov-Galerkin method is the symmetry of the stiffness matrix which is generally easy to solve, however more calculations are necessary to get the stiffness matrix. Whereas, in the Petrov-Galerking method, obtaining the stiffness matrix is easier but solving the system on other hand can be more difficult owing to its non-symmetry.

=Problem 3=

Given
Given Equation 1 on 8-1 which states $$\overset{w}{\rightarrow}.\overset{P(v)}{\rightarrow}=0$$
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$$\forall w\in R^{n}$$

 *
$$\displaystyle (Eq. 3.1) $$
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and $$\overset{w}{\rightarrow}.\overset{P(v)}{\rightarrow}=0$$ where $$\overset{w}{\rightarrow}=\sum \beta _{i}.a_{i}$$
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$$\forall (\beta _{1},\beta _{2}..\beta _{n})\in R^{n}$$

 *
$$\displaystyle (Eq. 3.2) $$
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 * }

Find
Show that $$\displaystyle (Eq. 3.1) $$ is equivalent to $$\displaystyle (Eq. 3.2) $$

Solution
For avoiding confusion we are designating the weight function as $$\overset{W}{\rightarrow}$$ in $$\displaystyle (Eq. 3.1)$$ and the weight function as $$\overset{w}{\rightarrow}$$ in $$\displaystyle (Eq. 3.2)$$.

Considering $$\displaystyle (Eq. 3.2) $$ and Substituting for $$\overset{W}{\rightarrow}$$

$$\sum \beta _{i}a_{i}.\overset{P(v)}{\rightarrow}=0$$
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$$\forall (\beta _{1},\beta _{2}..\beta _{n})\in R^{n}$$

 *
$$\displaystyle (Eq. 3.3) $$
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 * }

As $$a_{i}$$ is an orthonormal basis, performing a successive product by $$a_{i}$$ in the $$\displaystyle (Eq. 3.3) $$, we get

$$\begin{bmatrix} 1 &0 &0  &....0 \end{bmatrix}\beta _{1}.\overset{P(v)}{\rightarrow}=0$$

In the similar way

$$\begin{bmatrix} 0 &1 &0  &....0 \end{bmatrix}\beta _{2}.\overset{P(v)}{\rightarrow}=0$$

.

.

.

$$\begin{bmatrix} 0 &0 &0  &....1 \end{bmatrix}\beta _{n}.\overset{P(v)}{\rightarrow}=0$$;

Furthermore all the equations can be combined and written as
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$$ \begin{bmatrix} 1 &0 &...0 \\ 0 &1  &...0 \\ 0 &0  &...1 \end{bmatrix}\bigl(\begin{smallmatrix} \beta_{1}\\ .\\ .\\ \beta_{n} \end{smallmatrix}\bigr).\overset{P(v)}{\rightarrow}=\bigl(\begin{smallmatrix} 0\\ .\\ .\\ 0 \end{smallmatrix}\bigr) ... \forall (\beta _{1},\beta _{2}..\beta _{n})\in R^{n}$$

 *
$$\displaystyle (Eq. 3.4) $$
 * }
 * }

We can cleraly observe that the Matrix on the left hand side is the identity matrix $$\begin{bmatrix} I \end{bmatrix}$$; Moreover we have also written $$(\beta _{1},\beta _{2}..\beta _{n})$$ as a vector which further can be called as $$\overset{\beta}{\rightarrow}$$. and $$\displaystyle (Eq. 3.4) $$ becomes
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$$\overset{\beta}{\rightarrow}.\overset{P(v)}{\rightarrow}=0; \forall \overset{\beta}{\rightarrow} \in R^{n}$$
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 *
$$\displaystyle (Eq. 3.5) $$
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Which is identical to $$\displaystyle Eq. 3.1 $$.

=Problem 4= Show that
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$$ \int \frac{x^2}{1+x} dx = \frac{x^2}{2} - x + ln \left(1+x \right) + k $$
 *
$$\displaystyle (Eq. 4.1) $$
 * }
 * }

Find
From Lecture 9

4.1$$ \int lnxdx = xlnx -x $$

4.2$$ \int xlnxdx = \frac{1}{2} x^2 \left[lnx - \frac{1}{2} \right] $$

4.3$$ \int \frac{x^2}{1+cx} dx $$

4.4$$ \int \frac{x^2}{a+bx} dx $$

From Lecture 10

4.5 Find the exact solution $$ u \left(x \right) $$ for the problem (3) on 9-2

4.6 Plot $$ u \left(x \right) $$

4.1
Prove that
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$$ \int lnxdx = xlnx -x $$
 *
$$\displaystyle (Eq. 4.2) $$
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 * }

In order to integrate by parts we choose $$ f \left(x \right) = lnx $$ and $$ g' \left(x \right) = 1 $$

Now we apply the formulae for integration by parts which is:

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$$ \int f \left( x \right) g' \left( x \right)dx = f \left(x \right) g \left(x \right) - \int g \left(x \right) f' \left(x \right)dx $$
 *
$$\displaystyle (Eq. 4.3) $$
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 * }

Where $$ f' \left(x \right) = \frac{1}{x}, g \left(x \right) = x $$

Substituting these values into Eq.4.3 we get:
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$$ \int lnxdx = xlnx - \int \frac{x}{x} dx = xlnx -x $$
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 *
$$\displaystyle (Eq. 4.4) $$
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 * }

4.2
Prove that
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$$ \int xlnxdx = \frac{1}{2} x^2 \left[lnx - \frac{1}{2} \right] $$
 *
$$\displaystyle (Eq. 4.5) $$
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 * }

In order to integrate by parts we choose: $$ f \left(x \right) = lnx $$ and $$ g' \left(x \right) = x $$

Where $$ f' \left(x \right) = \frac{1}{x}, g \left(x \right) = \frac{x^2}{2} $$

Substituting these values into Eq.4.3 we get:

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$$ \int xlnxdx = \frac{x^2}{2} lnx - \int \frac{x^2}{2} \frac{1}{x} dx = \frac{x^2}{2} lnx - \int \frac{x}{2}dx = \frac{x^2}{2} lnx - \frac{x^2}{4} $$
 *
$$\displaystyle (Eq. 4.6) $$
 * }
 * }

Factoring out $$ \frac{1}{2} x^2 $$ we get:
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$$ \int xlnxdx = \frac{1}{2} x^2 \left[lnx - \frac{1}{2} \right] $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
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 *
$$\displaystyle (Eq. 4.5) $$
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 * }

4.3
Find
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$$ \int \frac{x^2}{1+cx} dx $$
 *
$$\displaystyle (Eq. 4.7) $$
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 * }

The integral can also be written as follows:

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$$ \int \frac{x^2}{1+cx} dx = \frac{1}{c} \int \left[x - \frac{x}{1+cx} \right]dx = \frac{1}{c} \left[\int xdx - \int \frac{x}{1+cx} dx \right] $$
 *
$$\displaystyle (Eq. 4.8) $$
 * }
 * }

To solve this integral, we made the next change of variables:
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$$ y = 1+cx \Rightarrow  x = \frac{y-1}{c}  \Rightarrow  dy = cdx $$
 *
$$\displaystyle (Eq. 4.9) $$
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 * }

Substituting these values into Eq.4.8 we get:
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$$ \int \frac{ \left(y -1 \right)^2}{c^2 y} \frac{dy}{c} $$
 *
$$\displaystyle (Eq. 4.10) $$
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 * }

Which can be expanded to:
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$$ \frac{1}{c^3} \int \frac{y^2 - 2y +1}{y} dy = \frac{1}{c^3} \int \left(y -2+ \frac{1}{y} \right) dy = \frac{1}{c^3} \left( \frac{y^2}{2} -2y+lny \right) $$
 *
$$\displaystyle (Eq. 4.11) $$
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Finally, replacing y as a function of x we have:
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$$ \frac{1}{c^3} \left[ \frac{ \left(1+cx \right)^2}{2} - 2 \left(1+cx \right) + ln \left(1+cx \right) \right] $$
 *
$$\displaystyle (Eq. 4.12) $$
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 * }

Choosing $$ c = 1 $$
 * {| style="width:100%" border="0"

$$ \int \frac{x^2}{1+x} dx = \frac{x^2}{2} - x - \frac{3}{2} + ln \left(1+x \right) $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 *
$$\displaystyle (Eq. 4.13) $$
 * style= |
 * }

 * Wolfram Alpha Proof for Part 3

4.4
Find
 * {| style="width:100%" border="0"

$$ \int \frac{x^2}{a+bx} dx $$
 *
$$\displaystyle (Eq. 4.14) $$
 * }
 * }

We start by changing the variables with:

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

$$ y = a +bx \Rightarrow  x = \frac{y - a}{b}  \Rightarrow  dx = \frac{dy}{b}$$
 *
$$\displaystyle (Eq. 4.15) $$
 * }
 * }

Replacing these values into Eq.4.14 we get:

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

$$ \int \frac{ \left( y-a \right)^2}{b^2 y} \frac{dy}{b} = \frac{1}{b^3} \int \left(y -2a+ \frac{a^2}{y} \right) dy = \frac{1}{b^3} \left( \frac{y^2}{2} - 2ay+ a^2 ln y \right) $$
 *
$$\displaystyle (Eq. 4.16) $$
 * }
 * }

Changing the variables we get:

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

$$ \frac{1}{b^3} \left( \frac{ \left(a+bx \right)^2}{2} - 2a \left( a+bx \right) + a^2 ln \left(a+bx \right) \right) $$
 *
$$\displaystyle (Eq. 4.17) $$
 * }
 * }

Choosing $$ a = b =1 $$ we have:
 * {| style="width:100%" border="0"

$$ \int \frac{x^2}{1+x} dx = \frac{ \left(1+x \right)^2}{2} - 2 \left(1+x \right) + ln \left(1+x \right) $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 *
$$\displaystyle (Eq. 4.18) $$
 * style= |
 * }

The result is the same as Eq.4.13

 * Wolfram Alpha Proof for Part 4

4.5
The problem is solving the differential equation:

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

$$ \frac{d}{dx} \left[ \left( 2+3x \right) \frac{du}{dx} \right] +5x = 0  \forall x \in \left]0,1 \right[  $$
 *
$$\displaystyle (Eq. 4.19) $$
 * }
 * }

With the boundary conditions:

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

$$ u \left(1 \right) = 4 ~,~ - \frac{du}{dx} \left(x = 0 \right) = 6 $$
 *
$$\displaystyle (Eq. 4.20) $$
 * }
 * }

The differential equation can be written:
 * {| style="width:100%" border="0"

$$ \frac{d}{dx} \left[ \left( 2+3x \right) \frac{du}{dx} \right] = -5x $$
 *
$$\displaystyle (Eq. 4.21) $$
 * }
 * }

Now integrate:

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

$$ \left[ \left( 2+3x \right) \frac{du}{dx} \right] = -5 \frac{x^2}{2} + k $$
 *
$$\displaystyle (Eq. 4.22) $$
 * }
 * }

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

$$ \frac{du}{dx}= -2.5 \frac{x^2}{\left(2 +3x \right)} + \frac{k}{\left( 2+3x \right)} $$
 *
$$\displaystyle (Eq. 4.23) $$
 * }
 * }

Now the solution of $$ u \left( x \right) $$ is:

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

$$ u \left( x \right) = -2.5 \int \frac{x^2}{ \left(2+3x \right)} dx + k \int \frac{dx}{ \left(2+3x \right)} $$
 *
$$\displaystyle (Eq. 4.24) $$
 * }
 * }

The first integral can be solved using Eq.4.17 with  $$  a = 2 ~,~ b=3 $$   and the second integral can be solved

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

$$ \int \frac{dx}{ \left(2+3x \right)} = \frac{1}{3} ln \left(2+3x \right) + q $$
 *
$$\displaystyle (Eq. 4.25) $$
 * }
 * }

Therefore the total solution is:
 * {| style="width:100%" border="0"

$$ u \left(x \right) = -2.5 \frac{1}{3^3} \left[ \frac{ \left(2+ 3x \right)^2}{2} - 2 \times 2 \left(2+3x \right) + 2^2 ln \left( 2+3x \right) \right] + \frac{k}{3} ln \left(2+3x \right) +q $$
 *
$$\displaystyle (Eq. 4.26) $$
 * }
 * }

After simplifying everything we obtain:

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

$$ u \left(x \right) = \frac{-5}{12}x^2 + \frac{30}{54}x + pln \left(2+3x \right) + r $$
 *
$$\displaystyle (Eq. 4.27) $$
 * }
 * }

Where $$ p $$ and $$ r $$ are constants and can be solved using the boundary conditions (Eq.4.20)

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

$$ u \left(1 \right) = 4 = \frac{-5}{12} + \frac{30}{54} + pln \left(5 \right) + r $$
 *
$$\displaystyle (Eq. 4.28) $$
 * }
 * }

The derivative of u is:

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

$$ \frac{du}{dx} = \frac{-10}{12}x + \frac{30}{54} + \frac{3p}{\left(2+3x \right)} $$
 *
$$\displaystyle (Eq. 4.29) $$
 * }
 * }

Therefore:

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

$$ \frac{du}{dx} \left(x=0 \right) = \frac{30}{54} + \frac{3p}{2} = -6  $$
 *
$$\displaystyle (Eq. 4.30) $$
 * }
 * }

Solving for $$ p $$ then gives:
 * {| style="width:100%" border="0"

$$ p = - \frac{118}{27} $$
 *
$$\displaystyle (Eq. 4.31) $$
 * }
 * }

Replacing $$ p $$ into Eq.4.28 gives:

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

$$\displaystyle r = 10.895 $$
 *
$$\displaystyle (Eq. 4.32) $$
 * }
 * }

The complete solution is:
 * {| style="width:100%" border="0"

$$ u \left(x \right) = \frac{-5}{12}x^2 + \frac{30}{54}x - \frac{118}{27} ln \left(2+3x \right) + 10.895  $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 *
$$\displaystyle (Eq. 4.33) $$
 * style= |
 * }

4.6
Plot $$ u \left( x \right) $$

=Problem 5=

Given

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

$$ \overset{b _{i}}{\rightarrow}  .\overset{P(v)}{\rightarrow}=0 ,  i=1,2,...,n $$

 *
$$\displaystyle (Eq. 5.1) $$
 * }
 * }

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

$$ \overset{w}{\rightarrow}  .\overset{P(v)}{\rightarrow}=0 $$ where $$  \overset{w}{\rightarrow} = \sum_{i=1}^{n}\alpha_{i} \overset{b _{i}}{\rightarrow} $$ ,
 *
$$\displaystyle (Eq. 5.2) $$
 * }
 * }

with $$ \forall (\alpha _{1},\alpha _{2},...,\alpha _{n})\in R^{n} $$

Solution
We have $$ \overset{b _{i}}{\rightarrow}  .\overset{P(v)}{\rightarrow}=0 $$

which means

$$ \overset{b _{1}}{\rightarrow}  .\overset{P(v)}{\rightarrow}=0 $$ , $$ \overset{b _{2}}{\rightarrow}  .\overset{P(v)}{\rightarrow}=0 $$ ,..., $$ \overset{b _{n}}{\rightarrow}  .\overset{P(v)}{\rightarrow}=0 $$.

So there exist arbitrary numbers $$ ( \alpha _{1}, \alpha _{2} ,..., \alpha _{n} ) $$ that

$$ \alpha _{1} \overset{b _{1}}{\rightarrow}  .\overset{P(v)}{\rightarrow}+ $$ $$ \alpha _{2} \overset{b _{2}}{\rightarrow}  .\overset{P(v)}{\rightarrow}+...+ $$ $$ \alpha _{n} \overset{b _{n}}{\rightarrow}  .\overset{P(v)}{\rightarrow}=0 $$

which is equivalent to the following:
 * {| style="width:100%" border="0"

$$ \overset{w}{\rightarrow}  .\overset{P(v)}{\rightarrow}=0 $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 *
$$\displaystyle (Eq. 5.3) $$
 * style= |
 * }

=Problem 6=

Given
Consider $$F= \left\{ 1, cos \left(iwx \right), sin \left(iwx \right)\right\} $$, where $$ i=1,2  $$ on interval $$  \left[0,T \right] $$,

i.e. $$\left\{1, cos \left(wx \right), cos \left(2wx \right), sin \left(wx \right), sin \left( 2wx\right) \right\} $$

Find
6.1 Construct $$\overset{\Gamma}{\rightarrow}$$ (F); Observe the properties of $$\overset{\Gamma}{\rightarrow}$$

6.2 Find det $$\overset{\Gamma}{\rightarrow}$$ (F)

6.3 Conclude that F is orthogonal

6.1
Construct $$\overset{\Gamma}{\rightarrow} (F) = \int _{0}^{T} b_i b_j dx   ,   i,j = 1,2,3,4,5   $$

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

$$\overset{\Gamma}{\rightarrow} (F) = \int _{0}^{T} \begin{bmatrix} 1 &coswx &cos2wx &sinwx &sin2wx \\ coswx &coswxcoswx &coswxcos2wx &coswxsinwx &coswxsin2wx \\ cos2wx &cos2wxcoswx &cos2wxcos2wx &cos2wxsinwx &cos2wxsin2wx \\ sinwx &sinwxcoswx &sinwxcos2wx &sinwxsinwx &sinwxsin2wx \\ sin2wx &sin2wxcoswx &sin2wxcos2wx &sin2wxsinwx &sin2wxsin2wx \end{bmatrix} dx $$

 *
$$\displaystyle (Eq 6.1) $$
 * }
 * }

First we solve the integral:
 * {| style="width:100%" border="0"

$$ I = \int _{0}^{T} cos(\frac{2m\pi}{T}x) cos(\frac{2n\pi}{T}x) dx$$
 *
$$\displaystyle (Eq. 6.2) $$
 * }
 * }

 * with $$\displaystyle m,n=0,1,2 $$

Then we can write:
 * {| style="width:100%" border="0"

$$ I= \frac{1}{2} (I+b+I-b)     $$

 *
$$\displaystyle (Eq. 6.3) $$
 * }
 * }

where we add and subtract the quantity
 * {| style="width:100%" border="0"

$$ b= \int_0^T sin(\frac{2m\pi}{T}x) sin(\frac{2n\pi}{T}x) dx$$

 *
$$\displaystyle (Eq. 6.4) $$
 * }
 * }

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

$$ I= \frac{1}{2} \int_0^T (cos(\frac{2m\pi}{T}x) cos(\frac{2n\pi}{T}x) + sin(\frac{2m\pi}{T}x) sin(\frac{2n\pi}{T}x) + cos(\frac{2m\pi}{T}x) cos(\frac{2n\pi}{T}x) - sin(\frac{2m\pi}{T}x) sin(\frac{2n\pi}{T}x))dx $$

 *
$$\displaystyle (Eq. 6.5) $$
 * }
 * }

And the identity: $$ cos(A \pm B)=cosAcosB \mp sinAsinB  $$,   can be used to have the following result:
 * {| style="width:100%" border="0"

$$ I= \frac{1}{2} \int_0^T (cos(\frac{2(m-n)\pi}{T} x)+cos(\frac{2(m+n)\pi}{T} x) )dx $$

 *
$$\displaystyle (Eq. 6.6) $$
 * }
 * }

Performing the integration we have:
 * {| style="width:100%" border="0"

$$ I=\frac{1}{2} [ \frac{T}{2(m-n)\pi} sin(\frac{2(m-n)\pi}{T} x) + \frac{T}{2(m+n)\pi} sin(\frac{2(m+n)\pi}{T} x) ] \mid_0^T $$

 *
$$\displaystyle (Eq. 6.7) $$
 * }
 * }

The parameters 2(m-n) and 2(m+n) are integers because n and m are integers; therefore, when the sine is evaluated at T it becomes $$sink\pi$$ where k is integer and therefore $$sink\pi=0 $$. When the function is evaluated at 0, clearly the sine is 0 too. However when m=n, the parameter 2(m-n) becomes 0 and the first term of the last equation is indeterminate. But when m=n the integral can be calculated as:

$$ \int_0^T (cos(\frac{2m\pi}{T}x ))^2 dx$$

$$=\frac{1}{2} \int_0^T (1+cos(\frac{4m\pi}{T}x )) dx$$

$$=\frac{1}{2} [x+ \frac{T}{4m\pi} sin(\frac{4m\pi}{T}x )]\mid_0^T$$

$$=\frac{T}{2} $$

We conclude that the integral is $$\displaystyle T/2 $$ when $$\displaystyle m=n $$ and 0 otherwise.

Following identical procedure, the other integrals can be calculated to obtain:

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

$$\overset{\Gamma}{\rightarrow} (F) = \frac{T}{2} \begin{bmatrix} 2 &0 &0  &0  &0 \\ 0 &1  &0  &0  &0 \\ 0 &0  &1  &0  &0 \\ 0 &0  &0  &1  &0 \\ 0 &0  &0  &0  &1 \end{bmatrix}$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 *
$$\displaystyle (Eq. 6.8) $$
 * style= |
 * }

and it can be seen that $$\overset{\Gamma}{\rightarrow} (F)$$ is diagonal.

6.2
Because $$\overset{\Gamma}{\rightarrow} (F)$$ is diagonal, the determinant can be easily calculated as the following:
 * {| style="width:100%" border="0"

$$Det(\overset{\Gamma}{\rightarrow} (F)) = T \frac{T}{2}\frac{T}{2}\frac{T}{2}\frac{T}{2} = \frac{T^5}{16} $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 *
$$\displaystyle (Eq. 6.9) $$
 * style= |
 * }

6.3
Conclude that $$\overset{\Gamma}{\rightarrow} (F)$$ is orthogonal.

Although $$\overset{\Gamma}{\rightarrow} (F)$$ is diagonal, and orthogonal, it is not equal to the identity matrix, therefore it is not equal to $$\delta_ij$$.

=Problem 7= Consider the family $$F=\begin{bmatrix}x,x^2,x^3,x^4 \end{bmatrix} $$

Is $$ F $$ an othogonal family evaluated at $$\Omega=\begin{bmatrix}0,1 \end{bmatrix} $$?

From Lecture 10

Given

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

$$F=\begin{bmatrix}x,x^2,x^3,x^4 \end{bmatrix} $$

 *
$$\displaystyle (Eq. 7.1) $$
 * }
 * }

Find
Is $$ F $$ an othogonal family evaluated at $$\Omega=\begin{bmatrix}0,1 \end{bmatrix} $$?

Solution
Let's start by calculating $$\Gamma (F)$$

$$\Gamma(F)=\int_{0}^{1} \begin{bmatrix} 1,x,x^2,x^3,x^4\\ x,x^2,x^3,x^4,x^5\\ x^2,x^3,x^4,x^5,x^6\\ x^3,x^4,x^5,x^6,x^7\\ x^4,x^5,x^6,x^7,x^8\\ \end{bmatrix}dx $$

Performing the integration we obtain:
 * {| style="width:100%" border="0"

$$\Gamma(F)= \begin{bmatrix} x,\frac{x^2}{2},\frac{x^3}{3},\frac{x^4}{4},\frac{x^5}{5}\\ \frac{x^2}{2},\frac{x^3}{3},\frac{x^4}{4},\frac{x^5}{5},\frac{x^6}{6}\\ \frac{x^3}{3},\frac{x^4}{4},\frac{x^5}{5},\frac{x^6}{6},\frac{x^7}{7}\\ \frac{x^4}{4},\frac{x^5}{5},\frac{x^6}{6},\frac{x^7}{7},\frac{x^8}{8}\\ \frac{x^5}{5},\frac{x^6}{6},\frac{x^7}{7},\frac{x^8}{8},\frac{x^9}{9}\\ \end{bmatrix}$$

 *
$$\displaystyle (Eq. 7.2) $$
 * }
 * }

When equation 7.2 is evaluated at (0,1) we obtain:
 * {| style="width:100%" border="0"

$$\Gamma(F)= \begin{bmatrix} 1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5}\\ \frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6}\\ \frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6},\frac{1}{7}\\ \frac{1}{4},\frac{1}{5},\frac{1}{6},\frac{1}{7},\frac{1}{8}\\ \frac{1}{5},\frac{1}{6},\frac{1}{7},\frac{1}{8},\frac{1}{9}\\ \end{bmatrix}$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 *
$$\displaystyle (Eq. 7.3) $$
 * style= |
 * }

Where it can clearly be observed that $$\Gamma(F)$$ is not orthogonal

=Problem 8= Show that equation 1 and equation 2 on 10-4 are equivalent.

From Lecture 11

Given
Let's define our equations:

Equation 1 on 10-4

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

$$\int_{\Omega }{}w^{h}(x)P(U^{h}(x))dx=0;\forall w^{h}(x)$$
 *
$$\displaystyle (Eq.8.1) $$
 * }
 * }

and Equation 2 on 10-4

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

$${\int}_{\Omega}{} b_{i}(x)P(U^{h}(x))dx=0 $$
 *
$$\displaystyle (Eq.8.2) $$
 * }
 * }

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

$$w^{h}(x)=\sum_{i=1}^{n}c_{i}b_{i}(x)$$
 *
$$\displaystyle (Eq.8.3) $$
 * }
 * }

Find
Show that $$eq.8.1$$ and $$eq.8.2$$ are equivalent.

Solution
If we chose $$c_{i}$$ to be orthogonal, we have:

$$c_{1}=\begin{bmatrix}1,0,...,0 \end{bmatrix} $$

$$c_{2}=\begin{bmatrix}0,1,...,0 \end{bmatrix} $$

$$c_{3}=\begin{bmatrix}0,0,...,1 \end{bmatrix} $$

When we multiply $$c_{i}$$ times $$b_{i}$$ we observe that multiply by $$c$$ is equivalent to multiplying by the identity matrix, therefore replacing $$w^{f}(x)$$ in $$\displaystyle Eq. 8.1$$ we have:

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

$$\int_{\Omega}^{ }w^{h}(x)P(U^{h}(x))dx=\int_{\Omega}^{ }\sum_{i=1}^{n}c_{i}b_{i}(x)P(U^{h}(x))dx=\int_{\Omega}^{ }I.b.P(U^{h}(x))dx=\int_{\Omega}^{ }b_{i}P(U^{h}(x))dx$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 *
$$\displaystyle (Eq. 8.4) $$
 * style= |
 * }

=Problem 9=

Given
Solving a differential equation using Weight Residual Form. From lecture 12

The partial differential equation is:

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

$$ \frac{\partial}{\partial x}\left(2\frac{\partial u}{\partial x}\right) + 3 = 0 $$
 *
$$\displaystyle (Eq.9.1) $$
 * }
 * }

The boundary conditions are:

$$ u\left(1\right) = 0$$

$$\frac{du}{dx}\left(0\right) = -4 $$

As weighting funtion should be used:

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

$$ b_{i} = \cos\left ( ix+\phi \right ) $$
 *
$$\displaystyle (Eq.9.2) $$
 * }
 * }

We will use $$\phi = \frac{\pi}{4}$$

Find

 * 9.1 Find an approximate solution of the form $$U^{h}=\sum_{i=0}^{n}d_{i}b_{i}$$ for n=2
 * 9.2 Find two equations that enforce the boundary conditions
 * 9.3 Project the weight residues
 * 9.4 Display the equations in matrix form
 * 9.5 Solve for $$\displaystyle d $$
 * 9.6 Construct $$U^{h}$$ and plot $$U^{h}$$ and $$u$$
 * 9.7 Repeat 9.1 to 9.6 for n = 4 and n = 6

9.1 Find an approximate solution with n=2
When n = 2 we have:
 * {| style="width:100%" border="0"

$$ d_{i}=\begin{bmatrix} d0 & d1 & d2 \end{bmatrix} $$
 *
$$\displaystyle (Eq.9.3) $$
 * }
 * }

and

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

$$ b_{i}=\begin{bmatrix} \cos \left (  \frac{\pi}{4}\right )& \cos  \left ( x+ \frac{\pi}{4}\right ) & \cos  \left ( 2x+ \frac{\pi}{4}\right ) \end{bmatrix} $$
 *
$$\displaystyle (Eq.9.4) $$
 * }
 * }

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

$$ U^{h}=d0\cos \left (\frac{\pi }{4} \right )+d1\cos \left ( x+\frac{\pi }{4} \right )+d2\cos \left ( 2x+\frac{\pi }{4} \right ) $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 *
$$\displaystyle (Eq. 9.5) $$
 * style= |
 * }

9.2 Find two equations that enforce the boundary conditions
The boundary conditions are:
 * {| style="width:100%" border="0"

$$u(1)= 0 =d0\cos \left (\frac{\pi }{4} \right )+d1\cos \left ( 1+\frac{\pi }{4} \right )+d2\cos \left ( 2+\frac{\pi }{4} \right ) $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 *
$$\displaystyle (Eq. 9.6) $$
 * style= |
 * }

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

$$ \frac{du}{dx}(0)=-4 = -d1\sin \left ( \frac{\pi }{4} \right )-2d2\sin \left ( \frac{\pi }{4} \right ) $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

 *
$$\displaystyle (Eq. 9.7) $$
 * style= |
 * }

9.3 Project the weight residues
In this case changing the function u(x) by the approximated function $$U^{h}$$ in the PDE (Eq.9.1), we found that P(u) is:

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

$$ P\left ( U^{h} \right )=2\frac{\partial^2 U^{h}}{\partial x^2}+3=-2\left ( d1\cos \left (x+ \frac{\pi }{4} \right ) +4d2\cos \left (2x+ \frac{\pi }{4} \right )\right )+3 $$
 *
$$\displaystyle (Eq.9.8) $$
 * }
 * }

The equation 9.8 can be written as:

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

$$ P\left ( U^{h} \right )= -2\begin{bmatrix}0 & \cos \left (x+ \frac{\pi }{4} \right ) & 4\cos \left (2x+ \frac{\pi }{4} \right ) \end{bmatrix}\begin{bmatrix} d0\\ d1\\

d2\end{bmatrix}+3 $$
 *
$$\displaystyle (Eq.9.9) $$
 * }
 * }

Projecting the residue we have:

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

$$ \int_{a}^{b}b(x)P\left ( U^{h} \right )= 0 $$ After substitution of (Eq.9.4) and (Eq.9.9) in (Eq.9.10) the following equation is obtained
 *
$$\displaystyle (Eq. 9.10) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$ 2\int_{0}^{1}\begin{bmatrix} \cos\left ( \frac{\pi }{4} \right ) \\\cos \left (x+ \frac{\pi }{4} \right ) \\ \cos \left ( 2x+\frac{\pi }{4} \right )
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

\end{bmatrix}\begin{bmatrix} 0 &\cos \left (  x+\frac{\pi }{4}\right ) & 4\cos \left ( 2x+\frac{\pi }{4} \right )\end{bmatrix}dx\begin{bmatrix} d0\\ d1\\

d2\end{bmatrix}=3\int_{0}^{1} \begin{bmatrix} \cos \left ( \frac{\pi }{4} \right )\\ \cos\left ( x+\frac{\pi }{4} \right ) \\

\cos \left ( 2x+\frac{\pi }{4} \right )\end{bmatrix}dx $$

 *
$$\displaystyle (Eq. 9.11) $$
 * style= |
 * }

9.4 Display the equations in matrix form
Performing the product and then the integration indicated in (Eq.9.11) we have:
 * {| style="width:100%" border="0"

$$ \begin{bmatrix} 0 & \cos\left ( 1 \right )+\sin\left (1 \right )  -1 & 2\cos\left (2  \right )+2\sin \left (2  \right )-2 \\ 0 & -\frac{1}{2} -\frac{1}{2}\sin \left (1 \right )^{2} +\frac{1}{2}\cos \left (1  \right )^{2}& -\frac{4}{3}+4\sin \left ( 1 \right )+\frac{4}{3}\cos\left ( 3 \right ) \\ 0 & -\frac{1}{3}+\sin \left ( 1\right ) +\frac{1}{3}\cos\left ( 3 \right )& 3-\sin \left ( 2 \right )^{2}+\cos \left ( 2 \right )^{2} \end{bmatrix}\begin{bmatrix} d0\\ d1\\
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

d2\end{bmatrix}=\begin{bmatrix} 2.121\\ 0.809\\

-0.537\end{bmatrix} $$

 *
$$\displaystyle (Eq. 9.12) $$
 * style= |
 * }

The latter equation is clearly of the form $$\displaystyle Kd=F$$. We can observe that K is not symmetric.

9.5 Solve for $$ d $$
In (Eq.9.12) we have three equations; but, as we need to enforce the boundary conditions we take the only one of them and solve it together with (Eq.9.6) and (Eq.9.7). From these equations we now have the system:

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

$$ \begin{bmatrix} \cos\left (\frac{\pi }{4} \right )  &\cos\left (1+\frac{\pi }{4}  \right )   & \cos\left (2+\frac{\pi}{4}  \right ) \\ 0 &\sin \left (\frac{\pi}{4} \right )  &2\sin \left (\frac{\pi}{4}  \right ) \\ 0 & 0.3818 & -1.0137 \end{bmatrix}\begin{bmatrix} d0\\ d1\\

d2\end{bmatrix}=\begin{bmatrix} 0\\ 4\\

2.12\end{bmatrix}$$
 *
$$\displaystyle (Eq.9.13) $$
 * }
 * }

After solving the system in (Eq.9.13) we obtain d:
 * {| style="width:100%" border="0"

$$\begin{bmatrix} d0\\ d1\\
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

d2\end{bmatrix}=\begin{bmatrix} 1.71\\ 5.61\\ 0.02 \end{bmatrix}$$

 *
$$$$
 * style= |
 * }

9.6 Construct $$U^{h}$$ and plot $$U^{h}$$ and $$u$$
Replacing d in (Eq.9.5) we obtain the approximated solution:

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

$$ U^{h}=1.71\cos \left (\frac{\pi }{4} \right )+5.61\cos \left ( x+\frac{\pi }{4} \right )+0.02\cos \left ( 2x+\frac{\pi }{4} \right ) $$
 *
$$\displaystyle (Eq.9.14) $$
 * }
 * }

On the other hand, the exact solution of the PDE (Eq.9.1) with the given boundary condition is:

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

$$ u\left ( x \right )=-\frac{3}{4}x^{2}-4x+\frac{19}{4} $$
 *
$$\displaystyle (Eq.9.15) $$
 * }
 * }

In Fig.9.1 we show the exact and the approximated solution.

9.7 Repeat 9.1 to 9.6) for i = 4 and i = 6
Now we repeat the procedure using i = 4 and i=6 in Eq.9.4

As the procedure is the same shown before, we develop a routine using Matlab to perform the products, the integration and to solve the system of equations. The routine is show in the appendix.

Fig.9.2 and Fig.9.3 shown the numerical solution for i = 4 and i = 6 respectively. It can be clearly seen that as the number of terms used in (Eq.9.4) the agreement between the analytical and numerical solution is better.

Fig.9.4 Show the error at x = 0.5 as funtion of the number of terms used in Eq.9.4. Clearly the error tends to be 0 as the number of terms increases. The error shown in Fig.9.4 was calculated as the absolute value of the difference between the analytical and the numerical solution at x = 0.5

 Plots:

=Appendix=

 Matlab Code:

=Team Member's Contributions=

 * Abhijeet Bhalerao
 * James Roark
 * Cedric Adam
 * Harrison Sheffield
 * Yu Hao
 * Fernando Casanova

title: User:Eml5526.s11.team01/Hwk2


text:
TESOL/Using need

The word "need" has a literal, informational meaning, and besides that it also accomplishes several other communicative tasks.

Mechanics
Usually "need" is used as a normal verb to express necessity.

Modal verb
Need can be a modal verb. It is more common in British English.
 * You needn't come to that meeting.
 * Need I say why?
 * That is the only meeting you need attend. You need attend that meeting.

Passive infinitive
"Need" plus a verb with -ing has the same meaning as a passive infinitive.

Regret
Saying someone need not have done something suggests that there was no reason to do it, that it was a bad idea. It differs in that respect from "did not need to have done."

Including the word "never" adds emphasis: someone need never have done something.

Commands
Using "need" in the present tense is often used to talk about obligation in the future. In particular, it can be used to give commands indirectly.
 * You'll need to finish your homework.
 * You'll need to come to my office.

Permission
Using "need not (do something)" grants permission not to do something at that moment. In this case, "need" is a modal verb. In a case where the listener might assume that they are obliged to do something, "need not" explicitly removes that obligation. This differs from using "need" as a regular verb, as in "You do not need to do something."

Possibility
"Need not" may also be used to equal "might not be." That is, it means that the possibility is less than 100%.

 * I'm sure she went to the wrong place. - She need not have: maybe she is just running late.
 * That face must mean that he hates me. - He needn't feel that way about you. Maybe he just remembered something unpleasant.

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

title: TESOL/Using need


text:
User talk:AnandTupsakhare

Hi.. Anand Tupsakhare here. Welcome to my page..

Page under construction..

title: User talk:AnandTupsakhare


text:
User:Laureali/Calino

Calino Trial Page
Welcome ! This page is a trial for actual project cal4ino. Actual content will be here soon...

Blending together creativity and innovation......
"Creativity is thinking up new things. Innovation is doing new things" quotes Theodore Levitt, the renowned American economist. CAL4INO, funded by the Education, Audiovisual and Culture Executive Agency (EACEA), attempts to blend both the creativity and the innovation together by encouraging people to think of new things and doing them. CAL4INO will use an interdisciplinary systems approach to identify specific conditions and factors that enable or inhibit creative innovation. The socio-cultural dimensions and human aspects across multiple sectors, industries and organizations influencing the creative capabilities of individuals will be given careful consideration. More specifically, CAL4INO proposes to,

 * Develop training modules for creativity and innovation and measurement instruments to benchmark outcomes through pan-European pilot demonstrations in 6 countries
 * Conduct surveys in order to assess needs and impact of creativity competencies on innovation
 * Launch a peer-reviewed journal supported with Web 2.0 tools and social networks for sustainable organic growth.

Creating Knowledge Societies.......
''"Never doubt that a small group of thoughtful, committed people can change the world. Indeed. It is the only thing that ever has." -Margaret Mead '' CAL4INO believes in the collective intelligence of people functioning as teams to amass creative competencies in order to develop meaningful innovations at personal, professional, social and organizational levels. The main research focus of CAL4INO is the creation of ‘knowledge societies’ with universities as clusters and ecosystems for fostering creativity and innovation. CAL4INO in itself has a powerful ‘knowledge society’ with 6 prestigious educational institutions, 2 entrepreneurial associations and 1 publishing company. They are namely, Riga International School of Economics and Business Administration (Latvia), University of Piraeus Research Centre (Greece), University of Wuppertal (Germany), University of Cambridge (UK), Queens university (Belfast), Laurea UAS (Finland), COTEC (Portugal), Emerald publications (UK) & Scottish Institute for Enterprise (Glasgow). The creative leaders, innovative educators and the inspiring entrepreneurs in the CAL4INO super team will synergize their diverse perspectives, experiences and skills in design, technology and business to establish research and training interventions that will make sustainable creativity and innovation as part of the lifelong learning.

Viral networking and Social media perspectives...........
CAL4INO recognizes the impact of the ever growing social media in the contemporary world. It plans to utilise the social media, the social networks and the Web 2. 0 tools innovatively for various creative activities. The overall objective is to create an extended ecosystem dedicated to creativity, innovation and enterprise based on Web 2.0 enabled social networks capable of organic growth, viral multiplication and sustainability. Laurea UAS holds the helm of the social media responsibility and will study on the adaptation of social media and other web 2.0 tools in enhancing creativity in educational, entrepreneurial and organizational processes. Laurea will support the entrepreneurial networks with a package of Web 2.0 tools to enhance knowledge sharing and also enhance the CAL4INO communication and dissemination activities using Web 2.0 tools and social networks. Besides, a Web 2.0 platform will be used in CAL4INO to promote broader dialogue and social engagement. The CAL4INO network will be linked with associated partner websites to enable productive networking. Virtual participation through webinars, webcasts and live casts will be encouraged to increase international diversity and constructive feedback.

Remarkable deliverables ...........
CAL4INO applies action oriented research to achieve a number of deliverables that will have a long lasting impact in understanding and as well steering individual and organizational creativity towards innovation and entrepreneurship. To list a few,
 * 1) Development of training modules for creativity and innovation by leveraging internationally recognized educational programs and benchmarking outcomes through pan-European pilot demonstrations in 6 countries
 * 2) Development of research methodology using web based measurements to provide insights on the creative competencies in individuals and in turn finding ways to increase their levels of creativity and innovation
 * 3) Launch of a peer-reviewed journal on social media by Emerald publications, supported by social networks and
 * 4) Dissemination of handbooks and manuals on best practices in fostering creativity and innovation

Opportunities for involvement..........
Creative learning for innovation is an essential aspect of entrepreneurship. Michael E. Porter, the distinguished economist and strategist believes that innovation is the central issue in economic prosperity. Instilling creative and innovative culture becomes inevitable for accomplishments at local, national, EU and global levels. The committed team of CAL4INO will make a noble effort to generate the first statistical data of its kind based on research and training to support EU policy level discussions and recommendations on creativity, innovation and entrepreneurship. CAL4INO is open to everyone to join hands to explore and showcase their endeavors involving creativity, innovation, entrepreneurship and social media. Participation, collaboration, events and publications are welcomed to be associated with CAL4INO’s outcomes. CAL4INO will be in operation during 2010 to 2013, but its findings and deeds will linger longer to contribute positive and progressive effects on collective innovation towards national prosperity. Professor Joseph Hassid of University of Piraeus Research Centre in Greece, who is one of the creators behind CAL4INO quotes that ''“the shortage of conventional competitiveness boosting resources can be adequately compensated by superior creativity.  Small nations may, under certain circumstances, excel in this!” ''- We believe that CAL4INO will certainly live up to his hope and bring value to national innovations.

Today's Featured Project

Left and right boxes

 * Pairs of boxes
 * You can create a pair of boxes using RoundBoxLeft, RoundBoxRight, and LeftRightBoxClose.

Note

 * Browser dependent
 * Some browsers will not see rounded corners. Rather the corners will appear square.

Laurea UAS

title: User:Laureali/Calino


text:
User talk:Laureali

 Hello Laureali, 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! --Ottava Rima (talk) 18:33, 27 January 2011 (UTC)

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MaintenanceBot (discuss • contribs) 03:48, 11 November 2013 (UTC)

title: User talk:Laureali


text:
User:Egm6341.s11.team1.arm/HW2

Part 1

 Matlab code:

Part 2A (Taylor series)
From Homework 1.2 ,

To integrate,

Using the error definition given in Meeting 7-1 :

From Homework 1.2 ,

By the Integral Mean Value Theorem,

Finding the maximum and minimum of $$f(x)$$ on the interval and applying the Intermediate Value Theorem,

So the error will be bounded depending on $$n$$ in the following table

Using equation 4.5, the value for the integral for each respective $$n$$ are:

 Matlab code:

Part 2B (Trapezoidal Rule)
For the following parts, Wolfram Alpha was used to determine the "actual" value of $$I(x)$$ to be

The trapezoidal rule states:

Exercising this for a variety of $$n$$ and replacing the value of $$f(0)$$ whenever encountered with $$f(0)=1$$ (from HW1.1 ), the following table is generated:

 Matlab code:

Part 2C (Simpson's Rule)
Simpson's rule states:

Exercising this for a variety of $$n$$ and replacing the value of $$f(0)$$ whenever encountered with $$f(0)=1$$ (from HW1.1 ), the following table is generated:

 Matlab code:

Part 2D (Gauss-Legendre Quadrature)
The Gauss-Legendre Quadrature states:

where the values for $$x_i$$ and $$w_i$$ are the roots of the Legendre polynomial of order $$n$$ and their associated weights, respectively.

Exercising this for a variety of $$n$$ and replacing the value of $$f(0)$$ whenever encountered with $$f(0)=1$$ (from HW1.1 ), the following table is generated:

 Matlab code:

Problem 11
$$p_2$$ must be integrated from $$x_0$$ to $$x_2$$. To simplify, $$x_0$$ and $$x_2$$ can be set to -1 and 1, respectively, without loss of generality.

title: User:Egm6341.s11.team1.arm/HW2


text:
User:Egm6341.s11.team2.zou/Math Temp

= Original math environment templates =

Original math environment templates is given by prof as one of his HW guides. They display equations as follows:

Equations without box:

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

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

\lim_{x \to 0}\frac{x} = \lim_{x \to 0}\frac{1} = 1 $$     (1.1)
 *
 * }

and

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

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

f(x) = 1 $$     (1.2)
 *
 * }

Equations with box:

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

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

\lim_{x \to 0}\frac{x} = \lim_{x \to 0}\frac{1} = 1

$$     (1.3)
 *
 * }

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

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

f(x) = 1

$$     (1.4)
 *
 * }

= Proposed math environment templates =

Since other teams may use prof's templates, so we can use another set of templates to avoid commonness.

As suggested by Chris, we'd better display our equations along the center line of the page, making it more formal, similar to publications.

I don't know whether we should use invariant boxes or adjustable boxes since I've never written any paper, but I think equations with adjustable boxes looks more lovely when there're 2 more boxed equations together.

General Equations with Eq number lying in the precise center:
and

General Equations with Eq number lying in the precise center, plus arrow:
and

(DEPRECATED)Equations with Eq number and invariable box(constant width: 80% of page):
and

Equations with adjustable box:
and

Equations with adjustable box, Eq number and arrow:
and

The mechanism of the templates:
The math templates are actually tables. Changing border="0" to border="1" will demonstrate the mechanism of this set of templates:

and

As we can see the equation consists of three parts: the left box of putting arrows, the middle box of displaying equation body, and the right box of putting equation numbers. The width of each box is set to 10%, 80%, 10% respectivly ensuring that the change of content in each part won't interfere with the other and that the equation lies precisely in the center of page. Also note that there's a smaller box without length definition which is inside the middle box, that's why the colored box can be adjustable to equation length.

title: User:Egm6341.s11.team2.zou/Math Temp


title: File:Nm1.s11.team1.HW2.fig2.4.1.jpg


text:
User:Egm5526.s11.team4.zou

= Self Introduction =

Hi, I’m Xiangyu Zou, an international student from China, from Mechanical and Aerospace Department, pursuing Master’s degree. My study interests are broad, not particular in a chosen field since I’m just a would-be Master – I have no idea whether to write a thesis because I just don’t have an idea. But this semester I want to get myself some knowledge about calculation and mechanics. So I have chosen both NM1 and FE1 and Continuum Mechanics. I’ve taken prof’s course “Principle of Engineering Analysis 1” last semester so I know how to type Mediawiki and Latex. I’m inclined to use MS Word2007 + MathType + some Macros. I think it’s efficient and easy to learn to use. And I have fundamentals in solving ODEs. However my knowledge in FE and NM is very limited, and my coding skills are modest, my English is very poor too. So, I hope we can help each other and have a nice collaboration. By the way, my given name “Xiangyu” is very difficult to pronounce, so if you’re my teammates please just call me my last name “Zou” or anything except “Xiangyu”. Thank you!

= Useful Links =

Lecture Plan

Course Wiki

Team4 info

Math_Templates

= Homework =

Homewoek#1

Homewoek#2

Homewoek#3

Homewoek#4

Homewoek#5

Homewoek#6

Homewoek#7

Homewoek#6_NM

Homewoek#7_NM

title: User:Egm5526.s11.team4.zou


text:
User:Egm5526.s11.team4.zou/Math Temp

= Original math environment templates =

Original math environment templates is given by prof as one of his HW guides. They display equations as follows:

Equations without box:

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

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

\lim_{x \to 0}\frac{x} = \lim_{x \to 0}\frac{1} = 1 $$     (1.1)
 *
 * }

and

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

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

f(x) = 1 $$     (1.2)
 *
 * }

Equations with box:

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

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

\lim_{x \to 0}\frac{x} = \lim_{x \to 0}\frac{1} = 1

$$     (1.3)
 *
 * }

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

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

f(x) = 1

$$     (1.4)
 *
 * }

= Proposed math environment templates =

Since other teams may use prof's templates, so we can use another set of templates to avoid commonness.

As suggested by Chris, we'd better display our equations along the center line of the page, making it more formal, similar to publications.

I don't know whether we should use invariant boxes or adjustable boxes since I've never written any paper, but I think equations with adjustable boxes looks more lovely when there're 2 more boxed equations together.

General Equations with Eq number lying in the precise center:

and

Equations with Eq number and invariable box(constant width: 80% of page):

and

Equations with adjustable box:

and

Equations with adjustable box, Eq number and arrow:

{| style="width:100%" border="0" cellpadding="5" cellspacing="0"
 * style="width:10%" text-align=left| $$\displaystyle \Rightarrow$$
 * style="width:80%" |
 * style="width:80%" |

(1.11)
 * style="width:10%" |
 * }

and

{| style="width:100%" border="0" cellpadding="5" cellspacing="0"
 * style="width:10%" text-align=left| $$\displaystyle \Rightarrow$$
 * style="width:80%" |
 * style="width:80%" |

(1.12)
 * style="width:10%" |
 * }

title: User:Egm5526.s11.team4.zou/Math Temp


text:
Talk:Complex socio-ecological systems/Genetic algorithms

This week’s readings on genetic algorithms went far beyond my smallish realm of knowledge, but I was guerdoned by reading some of this material over again and realizing how incredible the potential for some of this research is. And while most of the texts are not very new, a casual perusal of the some recent titles in Google Scholar reveals that researchers in the past few years are using these concepts in work on developing subdivisions, project scheduling, and even job training. Perhaps not quite the sexiest venues for this research, or at least ones we would would associate with thinkers like Axelrod, Gell-Mann and John Holland. Still, it illustrates how some higher level computing can be applied to some of the most mundane tasks. Apart from Axelrod’s extensive description of the Prisoner’s Dilemma, and Gell-Mann’s exhortation that we not confuse complex adaptive systems with everything else, what I took away most from this week’s set of readings was the promise of this research evident in Holland’s SA piece. The mere idea that one can transfer nonlinear genetic complexity into binary numbers is, I suppose, not particularly new, especially for the technorati out there, but it is incredible to imagine that strings of bit information of varying type have the ability to encode everything from a gamete to multiple generations of a population. And even more complex still is the simulation of evolution and natural selection with something like 8,000 rules (as of 1992, at least—one can imagine that the processing power and know-how has increased since that time). From the standpoint of complexity and systems thinking, what I find both alluring and potentially dangerous about this line of research is that one can go from computing a single genome, to a series of 1’s and 0’s, to the development and troubleshooting of a jet engine, to modeling of an ecosystem, and then back again to the 1’s and 0’s. It seems that the analogies in these schematics are encoded in numeric criteria with endless variability, using incomplete parts (chromosomes or other “coarse” elements, as Gell-Mann calls them) to create larger wholes. (Or as Holland says, “The remarkable ability of genetic algorithms to focus their attention on the most promising parts of a solution space is a direct outcome of their ability to combine strings containing partial solutions.) Whither go these wholes? If there is no design evident, is there not greater risk? This, to me, speaks about what’s compelling about complex adaptive systems, connections to AI, and ultimately to sentience (http://illigal.org/2005/01/29/computer-sentience-is-possible-says-holland/): we want to design intelligent, complex systems—but do we want to live with the outcomes? Tegeticula

Similarly to Sam, I was allured by this week's readings elucidation of how much processes of evolution can be applied to technology, computer science and even decision-making processes by the rule of choice, combination, outcomes and learning or adaptation. I always like to think how much duality is infused in our lives and the enormous possibilities of binary systems. Thinking about genes, where the possible re-combinations of genetic codes are infinite, is both exciting and overwhelming. I like to think about applying microcosmos functioning to macrocosmos reality, such as when Holland refers to mutation, for example. Mutation would be the unpredictable playing its role, does not matter if it's good or bad. So, in Resilience science, the "surprise" plays an important role on unnanticipated events that can put the system under risk or take it to another stability state or regime shift. I like Hollands discussion on exploration and exploitation, thinking about how these ideas could apply to Resilience Science. Another example of microcosmos reflecting in macro phenomena is the idea of emergence or emergent properties, that are also shown through computer simulation. Axelrod states that agent-based modelling can help to make us understand emergent properties of bigger systems - there is something in common in the minds and choices of humans. Social norms emerge from social/cultural systems and sustain cooperation. People choose sides based on affinity. What might be the role of social norms in cultural resilience? I was comparing the idea of biological evolution with the idea of cultural evolution, or, to be more realistic, biocultural evolution (Gell-Man) - with the genetic algorythm combination and the learning processes that lead to adaptation. But is it adaptation always good? The graph in Axelrod shows that people will learn and will tend to reach the fitness landscape. How can we think about humanity evolution and learning, when the same processes that lead to adaptation also lead to destruction of our natural resource base?Simoneathayde 13:52, 28 January 2011 (UTC)Simone

SOME QUESTIONS: 1. What genetic algorythms can teach/offer to improve our understanding and management of socio-ecological systems? 2. What is the difference and consequences of applying inductive X deductive methods in scientific research? 3. What is parallelism in genetic algorythm functioning? What are some advantages of parallelism? 4. What is the rational-choice paradigm and why scientists have shifted to focus on adaptive behavior rather than rational choice? 5. What would be some applications of agent-based modelling in the study of resilience of socio-ecological systems? 6. Is adaptation always good? What are some examples of maladaptation resulting from anthropic manipulation of the environment?SimoneathaydeSimone


text:
User talk:Egm5526.s11.team4.zou

 Hello Egm5526.s11.team4.zou, 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.
 * Find out about research activities on Wikiversity.
 * Explore Wikiversity with the links to your left.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.
 * Follow Wikiversity on twitter (http://twiter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

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! --Ottava Rima (talk) 22:15, 27 January 2011 (UTC)

title: User talk:Egm5526.s11.team4.zou


text:
User:Ernie.cordell

I have two Bachelor of Arts degrees from Indiana State University in Terre Haute, Indiana (not the one in Indiana, Pennsylvania).

I am a Linux user since 1997. The distribution I currently use is RedHat. I currently dual-boot with Microsoft Windows whose operating systems I have used in every version and revision level as well as IBM OS/DOS/VS,OS/VS,VM/CMS and OS/MVS, CDC NOS3.1, Prime's Primos and Datacraft/Harris Corporation's Vulcan.

I believe Civility is the most important Wikiversity policy.

Subpages:
 * Interesting userspace pages
 * Other accounts
 * Reference Desk Archiving Notes

title: User:Ernie.cordell


text:
User:Eml5526.s11.team3.akj/Homework 1

 Homework 1

= Homework 1 = Formulation of Partial Differential Equation (PDE) for a 1- Dimensional Problem.

Refer to lecture notes 6.1

--

Problem Statement
1.1 To find the 1-Dimensional Partial Differential Equation(PDE) of an elastic bar of length L and varying cross-section, subjected to a body force or distributed force f(x,t) along the axis of the bar.

1.2 Now if the bar has a rectangular cross-section where the breadth is constant, perform the balance of forces to formulate the 1-D PDE for this case.

Solution 1.1
Consider an element of length $$d\left(x\right)$$ of the elastic bar, the free body diagram of the bar is shown as below.

In the above figure $$A\left(x\right)$$, $$ \sigma\left (x\right)$$ and  $$n\left(x\right)$$ represent the area, stress and normal to the plane on the left cross-section of the bar and

$$A\left(x+dx\right)$$, $$\sigma\left(x+dx\right)$$ and   $$n\left(x+dx\right)$$ represent the area, stress and normal to the plane on the right cross-section of the bar.

For the left cross-section $$n\left(x\right)=-1$$ and for the right cross-section $$n\left(x+dx\right)=1$$

For the element of the elastic bar shown, we have $$m\left(x\right)=\rho(x) A(x)dx$$

By using Newton's 2nd law of motion, we can have

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

|$$\displaystyle N(x+dx)-N(x)+f(x)dx= \rho(x) A(x)dx\frac{\partial^2u }{\partial t^2} $$ $$
 * $$\displaystyle (Eq. 1)

 * }
 * }

where $$N\left(x+dx\right)$$ is the net force due to the stress $$\sigma\left(x+dx\right)$$ which is given by

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

|$$\displaystyle N\left(x+dx\right)=\sigma(x+dx)A(x+dx)n(x+dx)$$ $$
 * $$\displaystyle (Eq. 2)

 * }
 * }

$$N\left(x\right)$$ is the net force due to the stress $$\sigma\left(x\right)$$ which is given by $$N\left(x\right)=\sigma(x)A(x)n(x)$$

Substituting for $$N\left(x+dx\right)$$,$$N\left(x\right)$$,$$n\left(x+dx\right)$$,$$n\left(x\right)$$, we get

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

|$$\displaystyle \sigma (x+dx)A(x+dx)-\sigma (x)A(x)+f(x,t)dx=\rho (x)A(x)dx\frac{\partial^2u }{\partial t^2} $$ $$
 * $$\displaystyle (Eq. 3)

 * }
 * }

Dividing the above equation by $$d\left(x\right)$$, we get

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

|$$\displaystyle \frac{\sigma (x+dx)A(x+dx)-\sigma (x)A(x)}{dx}+f(x,t)=\rho (x)A(x)\frac{\partial^2u }{\partial t^2}$$ $$
 * $$\displaystyle (Eq. 4)

 * }
 * }

Applying the limit as $${dx}{\to0}$$ in the above equation,by the definition of partial derivative, as we know that

$$ \lim_{dx\to0}\frac{f(x+dx)-f(x)}{dx}= \frac{\mathrm{d} f(x)}{\mathrm{d} x} $$

the first term on the LHS reduces to $$ \frac{\partial\left[\sigma (x)A(x)\right]}{\partial x} $$, the second term on the LHS remains the same as there is no $$d\left(x\right)$$ term. the RHS also remains the same.

Equation (4) reduces to

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

\frac{\partial\left[\sigma (x)A(x)\right]}{\partial x} +f(x,t)=\rho (x)A(x)\frac{\partial^2u }{\partial t^2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)

 * }
 * }

The stress $$ \sigma\left (x\right)$$ can be expressed in terms of elastic modulus $$ E\left (x\right)$$ and strain as $$\sigma (x)=E(x)\frac{\partial u}{\partial x}$$

Hence the PDE for 1 Dimensional elastic bar is

$$ \frac{\partial\left[E(x)A(x)\frac{\partial u}{\partial x}\right]}{\partial x}+f(x,t)= \rho (x)A(x)\frac{\partial^2u }{\partial t^2} $$ $$ $$
 *  $$ \displaystyle
 *  $$ \displaystyle (Eq. 6)

Solution 1.2
The Partial Differential Equation (PDE) of an elastic bar subjected to an axial load derived in the section (1.1) is

Now, a particular case is taken into consideration, where the bar has a rectangular cross-section. The breadth of the bar is a constant and is equal to 'b'. But the height of the bar varies along the length of the bar and hence it is function of x, which is 'h(x).' (For figure, refer page 6-1 of [[media:fe1.s11.mtg6.djvu|meeting 6]]).

Let us consider the inertial force at the center of the thin strip of the elastic bar. Therefore the mass at that location is given by,

Performing the balance of forces similar to part 1.1 and rearranging the terms we get

Applying the limit as $${\frac{dx}{2}\to0}$$ in the above equation,by the definition of partial derivative

$$\displaystyle \lim_{dx\to0}\frac{f(x+dx)-f(x)}{dx}= \frac{\mathrm{d} f(x)}{\mathrm{d} x} $$

we get,

By Hooke's Law, the stress $$ \sigma\left (x\right)$$ can be expressed in terms of elastic modulus $$ E\left (x\right)$$ and strain $$\frac{\partial u}{\partial x}$$ as

$$\displaystyle \sigma (x)=E(x)\frac{\partial u}{\partial x} $$

Hence the PDE for 1 Dimensional elastic bar is

$$ \frac{\partial \left[E(x)bh(x)\frac{\partial u}{\partial x}\right]}{\partial x}+f(x,t)= \rho (x)bh(x)\frac{\partial^2u }{\partial t^2}$$ $$
 *  $$ \displaystyle (Eq. 6)

$$\frac{\partial\left[E(x)h(x)\frac{\partial u}{\partial x}\right]}{\partial x}+\frac{1}{b}f(x,t)= \rho (x)h(x)\frac{\partial^2u }{\partial t^2}$$ $$
 *  $$ \displaystyle (Eq. 7)

Reference
1. Newton's 2nd law of motion 2. Partial derivative 3. Hooke's Law

Contributing Authors
EML5526.S11.Team3.risher 21:35, 24 January 2011 (UTC)

Eml5526.s11.team3.akj 16:50, 24 January 2011 (EST)

EML5526.S11.Team3.vnarayanan 16:52, 24 January 2011 (EST)

Eml5526.s11.team3.ushnish12 16:54, 24 January 2011 (EST)

Eml5526.s11.team3.hylon 16:56, 24 January 2011 (EST)

Eml5526.s11.team3.perry 11:57, 26 January 2011 (UTC)

Eml5526.s11.team3.sahin 09:27, 26 January 2011 (EST)

title: User:Eml5526.s11.team3.akj/Homework 1