text:
User:Eml5526.s11.team5/sub2

= HW 2.1 =

Problem
Derive the expression for the Differential Equation in the 1-D Heat Problem

Find
By taking infinitesimal slice of the bar (shown in red in Figure 1),$$ dx $$, develop expression for elastodynamic response in Heat Problem in 1-D

Solution
Heat is transferred in the form of Conduction, Convection and Thermal Radiation. In this problem we will do Heat Conduction only.

Here we have consider a small infinitesimal portion of wall of thickness dx as shown in figure.

Let,The Heat Flux is represented by $$ \quad q$$.

{| style="width:100%" border="0" Here $$ \quad \tilde{m}(x) =\quad m(x).c$$
 * style="width:95%" |
 * style="width:95%" |

--Lokeshdahiya 20:32, 2 February 2011 (UTC)

= HW 2.2 =

Given
1.
 * {| style="width:100%" border="0"

$$ \underline{a}_i \bullet  \underline{a}_j = \delta_{ij} $$
 * style="width:95%" |
 * style="width:95%" |

where $$ \delta_{ij} =

\begin{cases} i = j \rightarrow 1 \\ i \neq j \rightarrow 0 \end{cases}

$$ $$
 *  $$ \displaystyle (Eq. 2.2.1)
 * }

2.
 * {| style="width:100%" border="0"

$$ \underline{b}_j = \underline{b}_{jk} \underline{a}_k $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 2.2.2)
 * }

3.
 * {| style="width:100%" border="0"

$$ [b_{jk}] =
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2  & 6 \end{bmatrix}

$$ $$
 *  $$ \displaystyle
 * }

eg. $$ \underline{b}_2 = 2a_1 + (-1)a_2 + 3a_3 $$

Consider

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

$$ \underline{v} = 5a_1 - 7a_2 -4a_3 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 2.2.3)
 * }

Find
1. $$ det[\underline{b}_{jk}] $$

2. Find Gram matrix $$ \underline{\Gamma}(\underline{b}_1, \underline{b}_2, \underline{b}_3) = \underline{K} $$, $$ det(\underline{\Gamma}) $$

3. Find $$ \underline{F} = {F_i} = {\underline{b}_i \bullet \underline{v}_i} $$

4. Solve $$ \underline{K} \underline{d} = \underline{F} $$ for $$ \underline{d} = {v_i} $$

5. Use $$ \underline{w}_i \bullet \underline{P}(v) = 0 $$, to find $$ \bar{\underline{K}}  \underline{d} = \bar{\underline{F}}  $$.

What is $$ \bar{\underline{K}} $$ and $$ \bar{\underline{F}} \ $$ ? $$ \underline{d} = {v_j} $$

,where $$ \underline{w}_i \quad $$ is lineraly independent family of vectors (basis), e.g. $${ \underline{w}_i, i = 1, 2, 3,... n} \quad $$

6. Solve for $$ \underline{d} $$; compare to $$ \underline{d} $$ in part 4

7. Observe symmetry properties of $$ \underline{K} $$ and $$ \bar{\underline{K}} $$

Solution
1.

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

$$ det[b_{jk}] = det
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2  & 6 \end{bmatrix}

$$ $$
 *  $$ \displaystyle
 * }

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

$$ a_{11}*(a_{22}*a_{33} - a_{32}*a_{23}) - a_{12}*(a_{21}*a_{33} - a_{31}*a_{23}) + a_{13}*(a_{21}*a_{32} - a_{31}*a_{22}) \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 2.2.4)
 * }

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

$$ 1*(-1*6 - 2*3) - 1*(2*6 - 3*3) + 1*(2*2 - 3*(-1)) = 1*(-6-6)-1(12-9)+1*(4+3)= -12-3+7 = -8 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle
 * }

$$ \det[ b_{jk}] = -8 \quad $$

2.

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

$$ \underline{\Gamma} = (\underline{b}_i \bullet \underline{b}_j) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 2.2.5)
 * }

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

$$ \underline{b}_1 = 1a_1 + 1a_2 + 1a_3 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.2.6a)
 * }

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

$$ \underline{b}_2 = 2a_1 + (-1)a_2 + 3a_3 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.2.6b)
 * }

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

$$ \underline{b}_3 = 3a_1 + 2a_2 + 6a_3 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.2.6c)
 * }

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

$$ \underline{\Gamma}(b_{11}) = \underline{b}_1 \bullet \underline{b}_1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \underline{\Gamma}(b_{12}) = \underline{b}_1 \bullet \underline{b}_2 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \underline{\Gamma}(b_{13}) = \underline{b}_1 \bullet \underline{b}_3 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \underline{\Gamma}(b_{21}) = \underline{b}_2 \bullet \underline{b}_1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

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

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

$$ \underline{\Gamma}(b_{ij}) = \underline{b}_i \bullet \underline{b}_j $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.2.7)
 * }

also by definition of dot product,

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

$$ b_{21} = b_{21} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.2.8)
 * }

The $$ \underline{\Gamma} \quad $$ matrix is symmetric

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

$$ \underline{\Gamma} =
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{bmatrix}

$$ $$
 *
 $$ \displaystyle (Eq. 2.2.9)
 * }

Using matlab (code is attached below)

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

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

\underline{\Gamma} =

\begin{bmatrix} 3  &  4  &  11 \\     4   &  14  &  22  \\    11  &  22  &   49 \end{bmatrix}

$$ $$
 *
 $$ \displaystyle
 * }

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

$$ det(\underline{\Gamma}) = 64 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

3.

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

$$ \underline{F} = {\underline{b}_i \bullet \underline{v}_i} = [b_{jk}][v] =
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2  & 6 \end{bmatrix}

\begin{bmatrix} 5 \\ -7 \\ -4 \end{bmatrix}

$$ $$
 *
 $$ \displaystyle (Eq. 2.2.10)
 * }

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

$$ \underline{F} =
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} -6 \\ 5 \\ -23 \end{bmatrix}

$$ $$
 *
 $$ \displaystyle
 * }

4.

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

$$ \underline{\Gamma} \underline{d}_i = \underline{F} \Rightarrow [d_j] = [\Gamma]^T [F] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.2.11)
 * }

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

$$ [d_j] =
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} 3  &  4  &  11 \\     4   &  14  &  22  \\    11  &  22  &   49 \end{bmatrix}^T

\begin{bmatrix} -6 \\ 5 \\ -23 \end{bmatrix} $$ $$
 *
 $$ \displaystyle
 * }

$$ [d_j] =

\begin{bmatrix} 8.3750 \\ 5.6250 \\ -4.8750 \end{bmatrix}

$$

5. Consider $$ w_i * \underline{P(v)} = 0 $$ and assume $$ w_i = a_i \quad $$

Therefore

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

$$ a_i * \underline{P(v)} = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.2.12)
 * }

By definition of $$ \underline{P(v)} $$

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

$$ \underline{P(v)} = \sum_{j=1}^n \underline{a}_j v_j - \underline{v} = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.2.13)
 * }

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

$$ \Rightarrow \underline{a}_i * \sum_{j=1}^n \underline{a}_j v_j = \underline{a}_i \bullet \underline{v} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.2.14)
 * }

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

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

\begin{matrix} \sum_{j=1}^n & (\underline{a}_i \bullet \underline{a}_j) & v_j & = & a_i \bullet \underline{v}\\ & \Downarrow & \Downarrow & & \Downarrow \\ & \bar{\underline{K}_{ij}} & \underline{d} & & \bar{\underline{F}} \end{matrix}

$$ $$
 *
 $$ \displaystyle (Eq. 2.2.15)
 * }

From given $$ Eq. 2.2.2 |quad $$

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

$$ [a_{jk}] = [b_{jk}]^T[b_i] =
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} 1 & 2 & 3 \\ 1 & -1 & 2 \\ 1 & 3  & 6 \end{bmatrix}

\begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}

=

\begin{bmatrix} 1 & 2 & 3 \\ 1 & -1 & 2 \\ 1 & 3  & 6 \end{bmatrix}

=

\bar{\underline{K}}

$$ $$
 *
 $$ \displaystyle
 * }

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

$$ a_1 = 1b_1 + 2b_2 + 3b_3 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.2.16a)
 * }

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

$$ a_2 = 1b_1 + (-1)b_2 + 2b_3 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.2.16b)
 * }

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

$$ a_1 = 1b_1 + 3b_2 + 6b_3  \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.2.16c)
 * }

Following the same procedure as before

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

$$ \underline{\Gamma}_a =
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} =

\underline{\Gamma}_a =

\begin{bmatrix} 14 &   5  &  25  \\     5    &  6   &  10 \\    25  &  10  &   46 \end{bmatrix}

$$ $$
 *
 $$ \displaystyle (Eq. 2.2.17)
 * }

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

$$ det(\underline{\Gamma_a}) = 64 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Concerning $$ F_a \quad $$, noting the the functions are orthonormal, we may convert from one bases per say "a" to another bases "b" and vice verse by using identity matrix
 * {| style="width:100%" border="0"

$$ [I] =
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} 1 &   0  &  0  \\     0    &  1   &  0 \\    0  &  0  &   1 \end{bmatrix}

$$ $$
 *
 $$ \displaystyle (Eq. 2.2.18)
 * }

Therefore:

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

$$ [F_a] = [I][F_b] \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.2.19)
 * }

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

$$ [F_a] =
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} 1 &   0  &    0 \\     0    &  1   &  0 \\    0  &  0  &   1 \end{bmatrix}

\begin{bmatrix} 5 \\ -7 \\ -4 \end{bmatrix}

=

\begin{bmatrix} 5 \\ -7 \\ -4 \end{bmatrix}

$$ $$
 *
 $$ \displaystyle (Eq. 2.2.20)
 * }

6.

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

$$ \bar{\underline{K}} \underline{d}_j =  \bar{\underline{F}} \Rightarrow [d_{ja}] = [\bar{\underline{K}}]^T [\bar{\underline{F}}]  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.2.21)
 * }

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

$$ [d_{ja}] =
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} 1 & 2 & 3 \\ 1 & -1 & 2 \\ 1 & 3  & 6 \end{bmatrix}^{-1}

\begin{bmatrix} 5 \\ -7 \\ -4 \end{bmatrix}

=

\begin{bmatrix} 8.3750 \\ 5.6250 \\ -4.8750 \end{bmatrix}

$$ $$
 *
 $$ \displaystyle
 * }

From a plain observation it could be concluded that displacement $$ d_{ja} \quad $$ is the same.

7.

Both matrices $$ \bar{\underline{K}} \quad $$ and $$ \underline{K} \quad $$ were symmetric matrices, netherless neither diagonal nor the non-diagonal values were equal. Also it should be noted that these 2 matrices determinants came out to be the same.

= HW 2.3 = == 2.3.1 Background     ==

In defining basis  $$ \left \{\underline b_{i},i=1,...,n  \right \} $$   for   $$ \mathbb{R}^{n} $$,   $$ \left \{\underline b_{i} \right \} $$   is not necessarily orthonormal, where:

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

$$ \underline b_{i}\cdot \underline b_{j}\neq \delta _{ij} $$  &   $$ \delta _{ij}=\left\{\begin{matrix} 1\ for\ i=j\\ 0\ for\ i\neq j \end{matrix}\right. $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.1)
 * }

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

$$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Fig. 2.3.1)
 * }

A vector  $$ \textrm{\underline v} $$   within basis   $$ \left \{\underline b_{i} \right \} $$   may be defined as:

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

$$ \textrm{\underline v}=\sum_{i=1}^{n}\ \textrm{v}_{i}\underline b_{i} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.2)
 * }

or

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

$$ \textrm{\underline v}=\sum_{j=1}^{n}\ \textrm{v}_{j}\underline b_{j} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.3)
 * }

So if

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

$$ \textrm{\underline P (\underline v)}=\sum_{j=1}^{n}\ \textrm{v}_{j}\underline b_{j} - \textrm{\underline v}=\underline 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.4)
 * }

then

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

$$ \underline b_{i}\cdot\textrm{\underline P (\underline v)}=\underline b_{i}\cdot \sum_{j=1}^{n}\ \textrm{v}_{j}\underline b_{j} - \underline b_{i}\cdot \textrm{\underline v}=\underline 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.5)
 * }

And if

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

$$ \underline b_{i}\cdot \sum_{j=1}^{n}\ \textrm{v}_{j}\underline b_{j}=\sum_{j=1}^{n}(\underline b_{i}\cdot \underline b_{j})\textrm{v}_{j}=\underline b_{i}\cdot \textrm{\underline v} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.6)
 * }

then

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

$$ \underline b_{i}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ i=1,...,n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.7)
 * }

The following equation

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

$$ \sum_{j=1}^{n}(\underline b_{i}\cdot \underline b_{j})\textrm{v}_j=\underline b_{i}\cdot \textrm{\underline v} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.8)
 * }

can also be written as

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

$$ \left [ K_{ij} \right ]\left \{ \textrm {v}_j \right \}=\left \{ F_i \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.9)
 * }

where

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

$$ \left [ K_{ij} \right ]=(\underline b_i\cdot \underline b_j) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.10)
 * }

The equation above may also be written in the form:

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

$$ \underline K \ \underline d=\underline F $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.11)
 * }

$$ \underline K \ \underline d=\underline F $$ may also be obtained by multiplying $$ \underline P(\textrm v) $$   as defined above with any linear independent family of vectors (basis), such as   $$ \left \{ \underline w_i, i=1,...,n \right \} $$   for example. In such a case,

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

$$ \underline w_i\cdot \underline P (v)=0 \ \forall \ i=1,...,n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.12)
 * }

This equation is more general than

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

$$ \underline b_{i}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ i=1,...,n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.13)
 * }

as previously defined.

Considering now

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

$$ \underline w\cdot \underline P (\underline v)=0 \ \forall \ \underline w\in \mathbb{R}^{n} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.14)
 * }

and using our  $$ \left \{ \underline b_i \right \} $$   basis while setting

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

$$ \underline w = \sum_{i}\alpha_i\underline b_i $$  , $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.15)
 * }

our above equation becomes equivalent to two statements:

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

$$ \underline w \cdot \underline P(\textrm{ \underline v})=0 \ \forall \ \underline w=\sum_{i}\alpha _i \underline b_i \ \in \ \mathbb{R}^n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.16)
 * }

and

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

$$ \underline w \cdot \underline P(\textrm{ \underline v})=0 \ \forall \ \left \{ \alpha_1,..., \alpha_n \right \}\ \in \ \mathbb{R}^n $$  , such that    $$ \underline w=\sum_{i}\alpha _i\underline b_i $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.17)
 * }

To prove that this last equation is equivalent to

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

$$ \underline b_{i}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ i=1,...,n $$ , three choices are made. $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.18)
 * }

In Choice #1, where  $$ \alpha_1=1\quad$$, $$\alpha_2=...=\alpha_n=0 \quad$$  ,   $$ \left \{\alpha _1, ..., \alpha _n  \right \} $$   are arbitrary, i.e.    $$ \forall \ \left \{\alpha _1, ..., \alpha _n  \right \} $$

Then,

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

$$ \underline w=\sum_{i}\alpha _i\underline b_i=\sum_{i}\left \{ 1,0,...,0 \right \}\underline b_i=\underline b_1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.19)
 * }

and therefore

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

$$ \underline w\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.20)
 * }

and

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

$$ \underline b_1\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.21)
 * }

In Choice #2, where  $$ \alpha_2=1\quad$$, $$\alpha_1=0\quad$$, $$\alpha_3=...=\alpha_n=0 \quad$$  ,   $$ \left \{ \alpha _i \right \}=\left \{ 0,1,0,...,0 \right \} $$

Then,

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

$$ \underline w=\sum_{i}\alpha _i\underline b_i=\sum_{i}\left \{ 0,1,0,...,0 \right \}\underline b_i=\underline b_2 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.22)
 * }

and therefore

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

$$ \underline w\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.23)
 * }

and

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

$$ \underline b_2\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.24)
 * }

In Choice #3, where  $$ \alpha_n=1,\ \alpha_1=...=\alpha_{n-1}=0 $$   ,$$ \left \{ \alpha_i \right \}=\left \{ 0,..., 0,1 \right \} $$

Then,

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

$$ \underline w=\sum_{i}\alpha _i\underline b_i=\sum_{i}\left \{ 0,...,0,1 \right \}\underline b_i=\underline b_n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.25)
 * }

and therefore

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

$$ \underline w\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.26)
 * }

and

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

$$ \underline b_{n}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ \left \{ \alpha_1,...,\alpha_n \right \}\in \ \mathbb{R}^n $$  such that    $$ \underline w=\sum_{i}\alpha_i\underline b_i $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.27)
 * }

This end result is perfectly equivalent to:

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

$$ \underline b_{i}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ i=1,...,n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.28)
 * }

== 2.3.2 Solution      ==

Now we shall re-attempt the analysis above, yet redefine our basis to be   $$ \left \{\underline a_{i},i=1,...,n  \right \} $$  , perhaps where

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

$$ \underline a_{i}\cdot \underline a_{j}= \delta _{ij} $$ , where    $$ \delta _{ij}=\left\{\begin{matrix} 1\ for\ i=j\\ 0\ for\ i\neq j \end{matrix}\right. $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.29)
 * }

which makes it an orthonormal basis.

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

$$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Fig. 2.3.2)
 * }

As before, a vector  $$ \textrm {\underline v} $$   within basis   $$ \left \{\underline a_{i} \right \} $$   may be defined as:

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

$$ \textrm{\underline v}=\sum_{i=1}^{n}\ \textrm{v}_{i}\underline a_{i} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.30)
 * }

or

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

$$ \textrm{\underline v}=\sum_{j=1}^{n}\ \textrm{v}_{j}\underline a_{j} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.31)
 * }

So if

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

$$ \textrm{\underline P (\underline v)}=\sum_{j=1}^{n}\ \textrm{v}_{j}\underline a_{j} - \textrm{\underline v}=\underline 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.32)
 * }

Then

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

$$ \underline a_{i}\cdot\textrm{\underline P (\underline v)}=\underline a_{i}\cdot \sum_{j=1}^{n}\ \textrm{v}_{j}\underline a_{j} - \underline a_{i}\cdot \textrm{\underline v}=\underline 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.33)
 * }

And if

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

$$ \underline a_{i}\cdot \sum_{j=1}^{n}\ \textrm{v}_{j}\underline a_{j}=\underline a_{i}\cdot \textrm{\underline v}=\sum_{j=1}^{n}(\underline a_{i}\cdot \underline a_{j})\textrm{v}_{j}=\sum_{j=1}^{n}\ \delta_{ij}\cdot \textrm{v}_{j} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.34)
 * }

then

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

$$ \underline a_{i}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ i=1,...,n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.35)
 * }

Considering again

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

$$ \underline w\cdot \underline P (\textrm{\underline v})=0 \ \forall \ \underline w\in \mathbb{R}^{n} $$  , $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.36)
 * }

but this time setting

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

$$ \underline w = \sum_{i}\beta_i\underline a_i $$ , then the above equation is equivalent to: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.37)
 * }

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

$$ \underline w \cdot \underline P(\textrm{\underline v})=0 \ \forall \ \left \{ \beta_1,..., \beta_n \right \}\ \in \ \mathbb{R}^n $$  , such that $$ \underline w=\sum_{i}\beta_i\underline a_i $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.38)
 * }

To now prove that the above equation is equivalent to

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

$$ \underline a_{i}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ i=1,...,n $$ , three choices are made once again: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.39)
 * }

For Choice #1 where  $$ \beta_1=1 \quad$$, $$\beta_2=...=\beta_n=0 \quad$$  ,   $$ \left \{\beta _1, ..., \beta _n  \right \} $$   are arbitrary, i.e.   $$ \forall \ \left \{\beta _1, ..., \beta _n  \right \} $$

Then

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

$$ \underline w=\sum_{i}\beta _i\underline a_i=\sum_{i}\left \{ 1,0,...,0 \right \}\underline a_i=\underline a_1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.40)
 * }

and therefore,

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

$$ \underline w\cdot \underline P(\textrm{\underline v})=0 $$  , $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.41)
 * }

and so

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

$$ \underline a_1\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.42)
 * }

For Choice #2, where  $$ \beta_2=1 \quad $$, $$\beta_1=0 \quad$$, $$\beta_3=...=\beta_n=0 \quad$$  ,   $$ \left \{ \beta _i \right \}=\left \{ 0,1,0,...,0 \right \} $$

Therefore,

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

$$ \underline w=\sum_{i}\beta _i\underline a_i=\sum_{i}\left \{ 0,1,0,...,0 \right \}\underline a_i=\underline a_2 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.43)
 * }

and

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

$$ \underline w\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.44)
 * }

so therefore,

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

$$ \underline a_2\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.45)
 * }

In Choice #3, where  $$ \beta_n=1,\ \beta_1=...=\beta_{n-1}=0 $$  ,   $$ \left \{ \beta_i \right \}=\left \{ 0,..., 0,1 \right \} $$

So then

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

$$ \underline w=\sum_{i}\beta _i\underline a_i=\sum_{i}\left \{ 0,...,0,1 \right \}\underline a_i=\underline a_n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.46)
 * }

and accordingly,

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

$$ \underline w\cdot \underline P(\textrm{\underline v})=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.47)
 * }

So finally,

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

$$ \underline a_{n}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ \left \{ \beta_1,...,\beta_n \right \}\in \ \mathbb{R}^n $$ , such that   $$ \underline w=\sum_{i}\beta_i\underline a_i $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.48)
 * }

This end result is perfectly equivalent to:

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

$$ \underline a_{i}\cdot \textrm{\underline P(\underline v)}=0\ \forall \ i=1,...,n $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.49)
 * }

= HW 2.4 =

Given\Find
Show $$ \int_{} \frac{x^2}{1+x}dx = \frac{x^2}{2} - x +log(1+x) + k $$

Do the following problems to lead this result

1) Show $$\int{log(x) }dx = xlog|x| - x $$ 2) Show $$\int{x log(x)}dx = \frac{1}{2}x^2\left[log(x)-\frac{1}{2}\right]$$ 3) Find $$\int{\frac{x^2}{1+cx}}dx$$ 4) Find $$\int{\frac{x^2}{a+bx}}dx$$ 5) Find the exact solution of $$ \displaystyle u(x) $$ for, $$ \displaystyle            \frac{d}\left[ (2+3x)\frac\right] + 5x = 0 $$ 6) Plot $$ \displaystyle u(x) $$

1) Show $$\int{logx}dx = xlogx - x $$
Using Integration By Parts,

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

$$ \int{u}dv =vu-\int{u}dv $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.4.1)
 * }

Let

Entering values from $$ Eq. 2.4.2 $$ into $$ Eq. 2.4.1 $$ yields

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

$$ xlogx-\int 1dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \quad xlogx-x $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

2) Show $$\int{x logx}dx = \frac{1}{2}x^2\left[logx-\frac{1}{2}\right]$$
Using Integration by parts, let

Now entering values from $$ Eq. 2.4.3 $$ into $$ Eq. 2.4.1 $$ yields

3) Find $$\int \frac{x^2}{1+cx}$$
Using Integration by parts, let

4) Find $$\int \frac{x^2}{a+bx}$$
Using integration by parts, let

Now, entering values from $$ Eq. 2.4.10 $$ into $$ Eq. 2.4.1 $$ yields,

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

$$ \int \frac{x^{2}}{a+bx}dx =(\frac{log(a+bx)}{b})( x^{2}) - \int{ (\frac{log(a+bx)}{b})2x}dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.4.11)
 * }

Now, integration by parts must be done again. Let

The equation now takes the form

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

$$(\frac{log(a+bx)}{b})( x^{2}) - [ vu - \int{vdu}] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.4.13)
 * }

Entering the values from $$ Eq. 2.4.12 $$ into $$ Eq. 2.4.13 $$ yields,

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

$$\frac{x^2}{b} log(a+bx) - [ ( \frac{x}{b}log(a+bx) + \frac{a}{b^2}log(a+bx) - \frac{x}{b})( 2x) - \int{( \frac{x}{b}log(a+bx) + \frac{a}{b^2}log(a+bx) - \frac{x}{b})} {(2dx)}] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Distributing terms,

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

$$\frac{x^2}{b} log(a+bx) - \frac{2x^2}{b} log(a+bx) + \frac{2ax}{b^2}log(a+bx) + \frac{2x^2}{b} + \frac{2}{b}\int{xlog(a+bx)}dx + \frac{2a}{b^2}\int{log(a+bx)}dx - \frac{2}{b}\int{x}dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.4.14)
 * }

Solving each integral in $$ Eq. 2.4.14 $$ individually,

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

$$\frac{2}{b}\int{xlog(a+bx)}dx = \frac{x^2}{b}\log(a+bx)-\frac{a^2}{b^3}\log(a+bx)-\frac{x^2}{2b}+\frac{ax}{b^2} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \frac{2a}{b^2}\int{log(a+bx)}dx = \frac{2ax}{b^2}log(a+bx)+\frac{2a^2}{b^3}log(a+bx) - \frac{2ax}{b^2} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$\frac{2}{b}\int{x}dx = -\frac{x^2}{b} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle (Eq. 2.4.15)
 * }

Substituting $$ Eq. 2.4.15 $$ into $$ Eq. 2.4.14 $$ yields

Which simplifies to

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

$$\frac{x^2}{2b}-\frac{ax}{b^2} + \frac{a^2}{b^3}log(a+bx) + k_{constant} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle (Eq. 2.4.16)
 * }

5) Find the exact solution for

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

$$ \frac{d}\left[ (2+3x)\frac\right] + 5x = 0 $$  ;    $$  \forall x \in ] 0,1 [ $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

With the following Boundary Conditions:
 * {| style="width:100%" border="0"

Essential: $$ \quad u(1)=4$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.4.17)
 * }

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

Natural: $$ \quad -\frac{d}{dx} u(x=0) =6$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.4.18)
 * }

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

$$ \frac{d}{du} [(2+3x)\frac{du}{dx}]=-5x $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.4.19)
 * }

Integrate both sides of $$ Eq. 2.4.19 $$,

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

$$ \int d [(2+3x)\frac{du}{dx}] = \int-5x dx$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$(2+3x)\frac{du}{dx}=\frac{-5x^2}{2}+c_1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$\frac{du}{dx}=\frac{-5x^2}{2(2+3x)}+\frac{c_1}{2+3x}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.4.20)
 * }

Apply Natural B.C., $$ Eq. 2.4.18 $$, to find $$ c_1 $$,
 * {| style="width:100%" border="0"

$$-\frac{du}{dx}(x=0) =6 \implies -6= 0+ \frac{c_1}{2}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$\quad c_1=-12$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.4.21)
 * }

Enter value from $$ Eq. 2.4.21 $$ into $$ Eq. 2.4.20 $$

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

$$\frac{du}{dx}=\frac{-5x^2}{2(2+3x)}+\frac{-12}{2+3x} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$du=\frac{-5x^2}{2(2+3x)}dx+\frac{-12}{2+3x}dx$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.4.22)
 * }

Integrate both sides of $$ Eq. 2.4.22 $$

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

$$\int du=\int\frac{-5x^2}{2(2+3x)}dx+\int\frac{-12}{2+3x}dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$u=-\frac{5}{2}\int \frac{x^2}{2+3x}dx-12\int \frac{dx}{2+3x} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$u=\frac{-5}{2}[\frac{x^2}{6}-\frac{2x}{9}+\frac{4log(2+3x)}{27}]-12[\frac{log(2+3x)}{3}]+c_2 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$u=\frac{-5x^2}{12} +\frac{5x}{9}-\frac{118}{27}log(2+3x)+c_2$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.4.23)
 * }

Apply Essential B.C., $$ Eq. 2.4.17 $$, to find

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

$$u(1)=4 \implies 4=\frac{-5}{12} +\frac{5}{9}-\frac{118}{27}log(2+3x)+c_2$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$\quad c_2=10.895$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.4.24)
 * }

Enter the value from $$ Eq. 2.4.24 $$ into $$ Eq. 2.4.23 $$ to obtain exact solution,

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

$$u(x)=\frac{-5x^2}{12} +\frac{5x}{9}-\frac{118}{27}log(2+3x)+10.895$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

6) Plot $$ \quad u(x)$$

= HW 2.5 =

Given
$$\overset{b _{i}}{\rightarrow}\centerdot \overset{P}{\rightarrow}\overset{(V)}{\rightarrow}=0$$

Find
Show that $$\overset{b _{i}}{\rightarrow}\centerdot \overset{P}{\rightarrow}\overset{(V)}{\rightarrow}=0$$  where i = 1, 2,. . ., n is the same as $$\overset{W}{\rightarrow}\centerdot \overset{P}{\rightarrow}\overset{(V)}{\rightarrow}=0$$   such that     $$\overset{W}{\rightarrow}=\sum\limits_{i=1}^{n}$$

Solution
For i = 1, 2,. . ., n

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

$$\overset{b _{i}}{\rightarrow}\centerdot\overset{P}{\rightarrow}\overset{(V)}{\rightarrow}=0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.5.1)
 * }

Multiplying the arbitrary constant αi   on both sides ;

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

$${{\alpha }_{i}}(\overset{b _{i}}{\rightarrow}\centerdot\overset{P}{\rightarrow}\overset{(V)}{\rightarrow})=0$$ $$ From the properties of Dot product of a vector, $$c\left( \vec{u}\cdot \vec{v} \right)=c(\vec{u})\cdot \vec{v}=\vec{u}\cdot c(\vec{v})$$, where 'c' is an arbitrary constant.
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.5.2)
 * }

Therefore, equation (2.5.2) becomes,

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

$${{\alpha }_{i}}\overset{b _{i}}{\rightarrow}\centerdot \overset{P}{\rightarrow}\overset{(V)}{\rightarrow}=0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

For i = 1;

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

$${{\alpha }_{1}}\overset{b _{1}}{\rightarrow}\centerdot \overset{P}{\rightarrow}\overset{(V)}{\rightarrow}=0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq.1)
 * }

For i = 2;

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

 * style="width:95%" |
 * style="width:95%" |

$${{\alpha }_{2}}\overset{b _{2}}{\rightarrow}\centerdot \overset{P}{\rightarrow}\overset{(V)}{\rightarrow}=0$$ $$
 *
 $$ \displaystyle (Eq. 2)
 * }

Similarly, For all i = n;

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

$${{\alpha }_{n}}\overset{b _{n}}{\rightarrow}\centerdot \overset{P}{\rightarrow}\overset{(V)}{\rightarrow}=0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. n)
 * }

Adding the equations Eq. 1, Eq. 2 .... upto Eq. n, we get,

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

$${{\alpha }_{1}}\overset{b _{1}}{\rightarrow}\centerdot \overset{P}{\rightarrow}\overset{(V)}{\rightarrow}$$ + $${{\alpha }_{2}}\overset{b _{2}}{\rightarrow}\centerdot \overset{P}{\rightarrow}\overset{(V)}{\rightarrow}$$ +. . . . . . + $${{\alpha }_{n}}\overset{b _{n}}{\rightarrow}\centerdot \overset{P}{\rightarrow}\overset{(V)}{\rightarrow}=0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.5.3)
 * }

Using the Distributive property of dot Product of a vector, Eqn. 2.5.3 can be written as,

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

$$({{\alpha }_{1}}\overset{b _{1}}{\rightarrow} + {{\alpha }_{2}}\overset{b _{2}}{\rightarrow} + . . . . . . + {{\alpha }_{n}}\overset{b _{n}}{\rightarrow})\centerdot \overset{P}{\rightarrow}\overset{(V)}{\rightarrow}=0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.5.4)
 * }

The above equation can be written as,

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

$$\sum\limits_{i=1}^{n}\centerdot \overset{P}{\rightarrow}\overset{(V)}{\rightarrow} = 0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.5.5)
 * }

Since $$\sum\limits_{i=1}^{n} = \overset{W}{\rightarrow},$$ Eq. 2.5.5 becomes

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

$$\overset{W}{\rightarrow}\centerdot \overset{P}{\rightarrow}\overset{(V)}{\rightarrow}=0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.5.6)
 * }

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

 * style="width:95%" |
 * style="width:95%" |

$$\overset{W}{\rightarrow}\centerdot \overset{P}{\rightarrow}\overset{(V)}{\rightarrow}=0$$ $$
 *
 $$ \displaystyle
 * }

Vignesh Solai Rameshbabu--Eml5526.s11.team5.srv 21:09, 30 January 2011 (UTC)

= HW 2.6 =

Given

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

$$F=\left \{ 1,cos i\omega x, sin i\omega x \right \}, i = \left \{ 1, 2 \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

on the interval, $$ \Omega =[0,T] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

i.e.,$$F=\left \{ 1,cos \omega x, cos 2\omega x, sin \omega x, sin 2\omega x \right \} = \left \{b_1(x), b_2(x), b_3(x), b_4(x), b_5(x) \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Find

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

1)Construct $$ \Gamma_{5x5}(F)$$. Observe the properties of $$ \Gamma $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

2) Find the det $$\Gamma$$(F) $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

3) Conclude F is orthogonal Basis, i.e., $$ \Gamma_{ij} = \delta_{ij} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Trigonometric Properties Used
$$ \implies \int_{0}^{T} cos (n \omega x) = 0, \forall n = \left \{ 1, 2, 3, ....\right \}, \omega = \frac{2\pi}{T} $$

$$ \implies \int_{0}^{T} sin (n \omega x) = 0, \forall n = \left \{ 1, 2, 3, ....\right \}, \omega = \frac{2\pi}{T} $$

$$ sin(x)cos(y) = \left [sin(x+y) + sin(x-y) \right ]/2,$$

$$cos(x)sin(y) = \left [sin(x+y) - sin(x-y) \right]/2,$$

$$cos(x)cos(y) = \left [cos(x-y) + cos(x+y) \right]/2,$$

$$sin(x)sin(y) = \left [cos(x-y) - cos(x+y)\right]/2.$$

Solution
1)

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

The Gram Matrix is defined as follows:, $$\Gamma_{ij}=<b_i,b_j>=\int_\Omega b_i(x)b_j(x)\,dx$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

It Can be observed that $$\Gamma_{ij}=\Gamma_{ji}$$, i.e. The $$ \Gamma $$ Matrix is symmetric $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

[Since $$ \int_\Omega b_i(x)b_j(x)\,dx = \int_\Omega b_j(x)b_i(x)\,dx$$] $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

Now Let us calculate each element in the $$ \Gamma $$ Matrix: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{1,1} = \int_{0}^{T} dx = \left [ x \right ]_{0}^{T} = T $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{1,2} = \Gamma_{2,1} = \int_{0}^{T} cos \omega x dx = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{1,3} = \Gamma_{3,1} = \int_{0}^{T} cos 2\omega x dx = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{1,4} = \Gamma_{4,1} = \int_{0}^{T} sin \omega xdx = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{1,5} = \Gamma_{5,1} = \int_{0}^{T} sin 2\omega x dx = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{2,2} = \int_{0}^{T} cos^{2} \omega x dx = \int_{0}^{T} \frac {cos 2\omega x +1}{2} = \frac {T}{2} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{2,3} = \Gamma_{3,2} = \int_{0}^{T} cos \omega x cos 2\omega x dx = \frac {1}{2} \int_{0}^{T} \left [ cos \omega x + cos 3\omega x \right ] dx = 0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{2,4} = \Gamma_{4,2} = \int_{0}^{T} cos \omega x sin \omega x dx = \frac {1}{2} \int_{0}^{T} sin 2\omega x dx = 0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{2,5} = \Gamma_{5,2} = \int_{0}^{T} cos \omega x sin 2\omega x dx = \frac {1}{2} \int_{0}^{T} \left [ sin 3\omega x + sin \omega x \right ] dx = 0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{3,3} = \int_{0}^{T} cos^{2} \omega 2x dx = \frac {1}{2} \int_{0}^{T} \left [ cos 4\omega x + 1 \right ] dx = \frac {T}{2}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{3,4} = \Gamma_{4,3} = \int_{0}^{T} cos 2\omega x sin \omega x dx = \frac {1}{2} \int_{0}^{T} \left [ sin 3\omega x - sin \omega x \right ] dx = 0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{3,5} = \Gamma_{5,3} = \int_{0}^{T} cos 2\omega x sin 2\omega x dx = \frac {1}{2} \int_{0}^{T} sin 4\omega x dx = 0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{4,4} = \int_{0}^{T} sin^{2} \omega x dx = \frac {1}{2} \int_{0}^{T} \left [ 1 - cos 2\omega x \right ] dx = \frac{T}{2}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{4,5} = \Gamma_{5,4} = \int_{0}^{T} sin \omega x sin 2\omega x dx = \frac {1}{2} \int_{0}^{T} \left [ cos \omega x - cos 3\omega x \right ] dx = 0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{4,4} = \int_{0}^{T} sin^{2} 2\omega x dx = \frac {1}{2} \int_{0}^{T} \left [ 1 - cos 4\omega x \right ] dx = \frac{T}{2}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

a) \Gamma_{5x5} \left (F \right) = \begin{bmatrix} T & 0 & 0 & 0 & 0\\ 0 & \frac{T}{2} & 0 & 0 & 0\\ 0 & 0 & \frac{T}{2} & 0 & 0\\ 0 & 0 & 0 & \frac{T}{2} & 0\\ 0 & 0 & 0 & 0 & \frac{T}{2} \end{bmatrix} $$

 * b) It is observed that the $$ \Gamma $$ Matrix is:

 * $$ \rightarrow $$ Diagnol

 * $$ \rightarrow $$ Symmetric

2)

 * $$ \det \left ( \Gamma \right ) = \frac {T^{5}}{16} $$
 * Since $$ \det \left ( \Gamma \right ) \ne 0, F $$ is a linearly independent family of functions.

3)

 * It can be observed that $$ \forall i \ne j, \Gamma_{ij} = 0,$$ Hence $$ F $$ is Orthogonal Basis.

--Eml5526.s11.team5.vijay 19:01, 1 February 2011 (UTC)--

= HW 2.7 =

Given

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

Consider $$ F=\left \{ 1,x,x^2,x^3,x^4 \right \} $$ on the interval $$ \Omega =[0,1] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Find

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

1. Construct the $$ \Gamma(F)$$ matrix and observe its propeties.
 * style="width:95%" |
 * style="width:95%" |
 * }

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

 * 2. Find the determinant of $$ \Gamma(F)$$.
 * }

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

3. Conclude if F is an orthogonal family., i.e., $$ \Gamma_{ij} = \delta_{ij} $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Solution
1)

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

The Gram Matrix is defined as:
 * style="width:95%" |
 * style="width:95%" |

 * $$\Gamma_{ij}=<b_i,b_j>=\int_\Omega b_i(x)b_j(x)\,dx$$

$$
 *
 $$ \displaystyle (Eq. 2.7.1)
 * }

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

For the case where $$ F=\left \{ 1,x,x^2,x^3,x^4 \right \} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }
 * {| style="width:100%" border="0"

$$ b_1(x)=1 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ b_2(x)=x \quad $$ $$
 * style="width:90%" |
 * style="width:90%" |
 *
 $$ \displaystyle
 * }

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

$$  b_3(x)=x^2 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$  b_4(x)=x^3  \quad$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ b_5(x)=x^4 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

Now we need to calculate each element in the $$ \Gamma $$ Matrix using eqn. 1.1: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

Since $$ \int_\Omega b_i(x)b_j(x)\,dx = \int_\Omega b_j(x)b_i(x)\,dx$$,  $$\Gamma_{ij}$$ will be equivalent to $$\Gamma_{ji}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{1,1} = \int_{0}^{1} 1*1*dx = \left [ x\right ]_{0}^{1} = 1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{1,2} = \Gamma_{2,1} = \int_{0}^{1} 1*x*dx = \left [ x^2/2\right ]_{0}^{1} = 1/2 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{1,3} = \Gamma_{3,1} = \int_{0}^{1} 1*x^2*dx = \left [ x^3/3 \right ]_{0}^{1} = 1/3 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{1,4} = \Gamma_{4,1} = \int_{0}^{1} 1*x^3*dx = \left [ x^4/4 \right ]_{0}^{1} = 1/4 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{1,5} = \Gamma_{5,1} = \int_{0}^{1} 1*x^4*dx = \left [ x^5/5 \right ]_{0}^{1} = 1/5 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{2,2} = \int_{0}^{1} x*x*dx = \left [ x^3/3\right ]_{0}^{1} = 1/3 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{2,3} = \Gamma_{3,2} = \int_{0}^{1} x*x^2*dx = \left [ x^4/4\right ]_{0}^{1} = 1/4 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{3,3} = \int_{0}^{1} x^2*x^2*dx = \left [ x^5/5 \right ]_{0}^{1} = 1/5 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{3,4} = \Gamma_{4,3} = \int_{0}^{1} x^2*x^3*dx = \left [ x^6/6\right ]_{0}^{1} = 1/6 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{3,5} = \Gamma_{5,3} = \int_{0}^{1} x^2*x^4*dx = \left [ x^7/7\right ]_{0}^{1} = 1/7 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{4,4} = \int_{0}^{1} x^3*x^3*dx = \left [ x^7/7\right ]_{0}^{1} = 1/7 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{4,5} = \Gamma_{5,4} = \int_{0}^{1} x^3*x^4*dx = \left [ x^8/8\right ]_{0}^{1} = 1/8 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \Gamma_{5,5} = \Gamma_{5,5} = \int_{0}^{1} x^4*x^4*dx = \left [ x^9/9\right ]_{0}^{1} = 1/9 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

Assembling the matrix we get: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

\Gamma_ \left (F \right) = \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:100%" border="0"

The matrix is observed to only be symmetric $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

2)

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

Using the matlab function det(X) the determinant of $$\Gamma$$ can be found to be: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * $$ \det \left ( \Gamma \right ) = 3.7493*10^{-12}$$

3)
 * {| style="width:100%" border="0"

To conclude if $$\Gamma$$ is orthogonal we compare it to a matrix formed by the Kronecker Delta. $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

The Kronecker Delta is defined by: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$\delta_{ij} = \begin{cases} 1, & \mbox{ for i=j} \\ 0, & \mbox{ for i≠j} \end{cases} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

Thus the general form of a the matrix is: $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \begin{bmatrix} 1    & \cdots & 0      \\ \vdots & \ddots & \vdots \\ 0     & \cdots & 1 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$ $$
 *
 $$ \displaystyle
 * }

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

Thus the $$ \Gamma $$ Matrix is not orthogonal $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

--Eml5526.s11.team5.savery 19:49, 2 February 2011 (UTC)

= HW 2.8 =

Given

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

$$\int_{\Omega}w^{h}(x)P(u^{h}(x))dx = 0 \;\; \forall\; w^{h}(x)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$\int_{\Omega}b_{i}(x)P(u^{h}(x))dx = 0 \;\; for\;\; i = 1,2,...,n$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Find
1.

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

$$\int_{\Omega}w^{h}(x)P(u^{h}(x))dx = 0  \;\Rightarrow\;  \int_{\Omega}b_{i}(x)P(u^{h}(x))dx = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

2.

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

$$  \int_{\Omega}b_{i}(x)P(u^{h}(x))dx = 0     \;\Rightarrow\;  \int_{\Omega}w^{h}(x)P(u^{h}(x))dx = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Solution
Let

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

$$\int_{\Omega}w^{h}(x)P(u^{h}(x))dx = 0 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.8.1)
 * }

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

$$\int_{\Omega}b_{i}(x)P(u^{h}(x))dx = 0 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.8.2)
 * }

1)To prove

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

$$\int_{\Omega}w^{h}(x)P(u^{h}(x))dx = 0  \;\Rightarrow\;  \int_{\Omega}b_{i}(x)P(u^{h}(x))dx = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Choice 1:
 * $$\;\; \big[c_{1}, c_{2}, c_{3},.........., c_{n}\big] = \big[ 1,0,0,...,0 \big] $$

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

$$\therefore\;\; w^{h}(x) = b_{1}(x) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Therefore Eq. 2.8.1 becomes,

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

$$\displaystyle \;\;\int_{\Omega}b_{1}(x)P(u^{h}(x))dx = 0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.8.3)
 * }

Choice 2:
 * $$\;\; \big[c_{1}, c_{2}, c_{3},............, c_{n}\big] = \big[ 0,1,0,...,0 \big] $$

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

$$\therefore\;\; w^{h}(x) = b_{2}(x) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Therefore Eq. 2.8.1 becomes,

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

$$\displaystyle \;\;\int_{\Omega}b_{2}(x)P(u^{h}(x))dx = 0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.8.4)
 * }

Choice n:
 * $$\;\; \big[c_{1}, c_{2}, c_{3},............, c_{n}\big] = \big[ 0,0,0,...,1 \big] $$

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

$$\therefore\;\; w^{h}(x) = b_{n}(x) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Therefore Eq. 2.8.1 becomes,

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

$$\displaystyle \;\;\int_{\Omega}b_{n}(x)P(u^{h}(x))dx = 0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.8.5)
 * }

In General, Eq. 2.8.3, Eq. 2.8.4, Eq. 2.8.5 can be written as,

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

|$$\displaystyle \int_{\Omega}b_{i}(x)P(u^{h}(x))dx = 0 \;\; where\;\; i = 1,2,...,n  \;( which \;is \;the\; same \;as\; Eq. 2.8.2) $$ $$
 *
$$\displaystyle
 * }
 * }

2)To prove

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

$$  \int_{\Omega}b_{i}(x)P(u^{h}(x))dx = 0     \;\Rightarrow\;  \int_{\Omega}w^{h}(x)P(u^{h}(x))dx = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

For i = 1, 2,. . ., n

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

$$\int_{\Omega}b_{i}(x)P(u^{h}(x))dx = 0 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.8.6)
 * }

Multiplying the arbitrary constant $$\;\;c_{i}\;\;$$   on both sides ;

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

$$\;\int_{\Omega}c_{i}b_{i}(x)P(u^{h}(x))dx = 0 \;\; for\;\; i = 1,2,...,n$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.8.7)
 * }

For i = 1;

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

$$\;\int_{\Omega}c_{1}b_{1}(x)P(u^{h}(x))dx = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq.1)
 * }

For i = 2;

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

$$\;\int_{\Omega}c_{2}b_{2}(x)P(u^{h}(x))dx = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq.2)
 * }

Similarly, For all i = n;

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

$$\;\int_{\Omega}c_{n}b_{n}(x)P(u^{h}(x))dx = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq.n)
 * }

Adding the equations Eq. 1, Eq. 2 .... upto Eq. n, we get,

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

$$\int_{\Omega}c_{1}b_{1}(x)P(u^{h}(x))dx + \int_{\Omega}c_{2}b_{2}(x)P(u^{h}(x))dx + .\; .\;. + \int_{\Omega}c_{n}b_{n}(x)P(u^{h}(x))dx = 0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.8.8)
 * }

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

$$\;\;\int_{\Omega}\left [ c_{1}b_{1}(x) + c_{2}b_{2}(x)+ .\; .\;. + c_{n}b_{n}(x)\right ]P(u^{h}(x))dx = 0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.8.9)
 * }

The above equation can be written as,

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

$$\displaystyle \Rightarrow\;\;\int_{\Omega}\sum_{i=1}^{n}c_{i}b_{i}(x)P(u^{h}(x))dx = 0$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.8.10)
 * }

Since $$\sum\limits_{i=1}^{n} = \;w^{h}(x)$$ Eqn(2.8.10) becomes

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

$$\int_{\Omega}w^{h}(x)P(u^{h}(x))dx = 0 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.8.11)
 * }

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

 * style="width:95%" |
 * style="width:95%" |

$$\int_{\Omega}w^{h}(x)P(u^{h}(x))dx = 0 \;\; \forall\; w^{h}(x)\;\;\; (which \;is\; the\; same\; as\; Eq. 2.8.1)$$ $$
 *
 $$ \displaystyle
 * }

Vignesh Solai Rameshbabu--Eml5526.s11.team5.srv 21:09, 30 January 2011 (UTC)

= HW 2.9 =

Given

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

$$ \omega = ]0,1[ \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.1)
 * }

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

$$ a_2 = 2 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.2)
 * }

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

$$ f = 3 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.3)
 * }

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

$$ \frac{\partial{u}}{\partial{t}}(x,t) = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.4)
 * }

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

$$ \Gamma_g = 1 $$, $$ g = 0 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ u(x=1) = 0 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.5)
 * }

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

$$ \Gamma_h = 0 $$, $$ h = 4 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

 * style="width:95%" |
 * style="width:95%" |

$$ -\frac{du^h}{dx}(0) = 4 \quad $$ $$
 *
 $$ \displaystyle (Eq. 2.3.6)
 * }

Find
 Find approximate solution $$ u^n \quad $$ and compare it to exact one

Consider base function $$ b_j(x), j = 0, 1, 2, 3,... n \quad $$

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

$$ b_j(x) = cos(jx+\phi) \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.7)
 * }

Consider $$ \phi = \frac{\pi}{4} $$ and $$ \phi = \frac{\pi}{2} $$

(1) Let $$ n = 2 \Rightarrow ndof = n+1 = 3 \quad $$

Generate variables considered for $$ n = 2 \quad $$ case

(2) Find 2 eqs that enforce b.c.'s for $$ u^h(x) = \sum_{j -0}^{n} d_j b_j(x) $$

(3) Find 1 more eq. to solve for $$ \underline{d} = d_j (j = 0, 1, 2) $$ by projecting the residue $$ \underline{p}(u^h) $$ on a basis function $$ b_k(x) \quad $$ with $$ k = 0, 1, 2 \quad $$. The additional equation must be lineally independent from the above 2 equations in part (2).

(4) Display 3 equations in matrix form

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

$$ \underline{K}\underline{d} = \underline{F} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.8)
 * }

Observe if $$ \underline{K} $$ is symmetric

(5) Solve for $$ \underline{d} \quad $$

(6) Construct $$ u_n^k(x) $$ and plot $$ u_n^k(x) $$ vs. $$ ux) $$

(7) Repeat steps (1) to (6) for

 * (7.1) $$ n = 4 \quad $$
 * (7.2) $$ n = 6 \quad $$

(8) Compute $$ u_n^h (x = 0.5) \quad $$ for $$ n = 2, 4, 6 \quad $$

error

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

$$ e_n(0.5) = u_n(0.5) - u_n^h(0.5) \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.9)
 * }

Plot $$ e_n(0.5) \quad $$ vs. $$ n \quad $$

Solution
(1)

For $$ n = 2 \quad $$

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

$$ \underline{b} =
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} b_0     & b_1 & b_2 \end{bmatrix}

$$ $$
 *
 $$ \displaystyle (Eq. 2.3.10)
 * }

From base function expression $$ b_j(x) = cos(jx+\phi) \quad $$

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

$$ \underline{b} =
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} cos(0x+\phi)     & cos(1x+\phi) & cos(2x+\phi) \end{bmatrix}

=

\begin{bmatrix} cos(\phi)     & cos(x+\phi) & cos(2x+\phi) \end{bmatrix}

$$ $$
 *
 $$ \displaystyle (Eq. 2.3.11)
 * }

Unknown function

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

$$ \underline{d} =
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} d_0     & d_1 & d_2 \end{bmatrix} $$ $$
 *
 $$ \displaystyle (Eq. 2.3.12)
 * }

Approximate solution

(2)

at B.C. $$ u(1) = 0 \quad $$

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

$$ u^h = \sum_{i=0}^{n=2} d_ib_i $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.13)
 * }

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

$$ u^h(x) = d_0cos(\phi) + d_1cos(x+\phi) + d_2cos(2x+\phi) \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.14)
 * }

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

$$ u^h(1) = 0 = d_0cos(\phi) + d_1cos(1+\phi) + d_2cos(2+\phi) =
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} cos(\phi)     & cos(1+\phi) & cos(2+\phi) \end{bmatrix}

\begin{bmatrix} d_0     \\ d_1 \\ d_2 \end{bmatrix} $$ $$
 *
 $$ \displaystyle
 * }

at B.C $$ \frac{du}{dx}(0) = 4 $$

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

$$ (u')^h(x) = \frac{du}{dx} = \sum_{i=0}^{n=2} d_ib_i' $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.15)
 * }

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

$$ \underline{b}' =
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} 0     & -sin(x+ \phi) & -2sin(2x+\phi) \end{bmatrix}

$$ $$
 *
 $$ \displaystyle (Eq. 2.3.16)
 * }

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

$$ (u')^h(0) = - 4 = d_00 - d_1sin(\phi) - d_22sin(\phi) =
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} 0     & sin(\phi) & 2sin(\phi) \end{bmatrix}

\begin{bmatrix} d_0     \\ d_1 \\ d_2 \end{bmatrix} $$ $$
 *
 $$ \displaystyle
 * }

(3)

By projecting the residue part of a function on the bases, 3rd equation can be generated:

From the definition of elastodynamics case of ODE function $$ P_u(x) \quad $$

Rewriting $$ Eq. 2.3.7 $$

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

$$ b_j(x) = cos(jx+\phi) \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

In this, more specific, "self adjoin" case

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

$$ P(u^h) :=[a_2u']' + a_0u - g = 0 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.17)
 * }

noting

$$ g = 0 \Rightarrow \quad $$ no forcing function (steady state)

$$ a_0u = f = 3 \quad $$ (given)

$$ a_2 = 2 \quad $$ (given)

$$ P(u^h) :=[2u']' + 3 = 0 \Rightarrow \quad $$ since $$ u \quad $$ is given as constant 2 (not function of "x")

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

 * style="width:95%" |
 * style="width:95%" |

$$ P(u^h) :=2u'' + 3 = 0\quad = 2\frac{d^2u}{dx} +3 = 0 $$ $$
 *
 $$ \displaystyle (Eq. 2.3.18)
 * }

From $$ Eq. xxx $$, $$ u^h \quad $$ was defined as

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

$$ u^h = \sum_{i=0}^{n=2} d_ib_i = d_0cos(\phi) + d_1cos(x+\phi) + d_2cos(2x+\phi) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.19)
 * }

Taking $$ 2^{nd} \quad $$ derivative with respect to "x"

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

$$ u^{h''} = 0 - d_1cos(x+\phi) - 4d_2cos(2x+\phi) \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.20)
 * }

Therefore

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

$$ P(u^h) := 2(0 - d_1cos(x+\phi) - 4d_2cos(2x+\phi)) +3 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.21)
 * }

Afterwords, noting that

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

$$ \int_\omega w P(u^h) dx := 0 \Rightarrow w_i = b_i \Rightarrow \int_\omega b_i P(u^h) dx := 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.22)
 * }

which is also noted in $$ 10-4, Eq. (2) \quad $$

and substituting for $$ P(u^h) \quad $$ equation above, we get

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

 * style="width:95%" |
 * style="width:95%" |

$$ \int_\omega b_i ([2(0 - d_1cos(x+\phi) - 4d_2cos(2x+\phi))] +3 dx) := 0 \quad $$ $$
 *
 $$ \displaystyle (Eq. 2.3.23)
 * }

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

 * style="width:95%" |
 * style="width:95%" |

$$ \int_\omega b_i [2(0 - d_1cos(x+\phi) - 4d_2cos(2x+\phi))] dx := \int_\omega b_i -3 dx \quad $$ $$
 *
 $$ \displaystyle (Eq. 2.3.24)
 * }

recalling that $$ \underline{b} =

\begin{bmatrix} cos(\phi)  \\ cos(x+\phi) \\ cos(2x+\phi) \end{bmatrix} $$

and taking any of the equations as the last needed equation, for example $$ b_1 \quad $$, we get

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

$$ \int_\omega
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} cos(\phi)  \\ cos(x+\phi) \\ cos(2x+\phi) \end{bmatrix}

[2(0 - d_1cos(x+\phi) - 4d_2cos(2x+\phi))] dx := \int_\omega

\begin{bmatrix} cos(\phi)  \\ cos(x+\phi) \\ cos(2x+\phi) \end{bmatrix} -3 dx \quad $$ $$
 *
 $$ \displaystyle (Eq. 2.3.25)
 * }

Doing symbolic integration and evaluating interval on $$ \omega = ]0,1[ \quad $$ we get

The outputs are as follow:

$$ \int_\omega

\begin{bmatrix} cos(\phi)  \\ cos(x+\phi) \\ cos(2x+\phi) \end{bmatrix}

[2(0 - d_1cos(x+\phi) - 4d_2cos(2x+\phi))] dx := $$ 0   0.3818   -1.0137         0    0.2919    0.7126         0    0.1781    2.3464

$$ \int_\omega

\begin{bmatrix} cos(\phi)  \\ cos(x+\phi) \\ cos(2x+\phi) \end{bmatrix} -3 dx \quad := $$

2.1213   0.8099   -0.5376

(4)

Assembling the matrix $$ \underline{K} \underline{d} = \underline{F} \quad $$ from 2 B.C. constrains and 1 more equation generated by weighting factor method:

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

$$ \begin{bmatrix} cos(\phi)     & cos(x+\phi) & cos(2x+\phi) \\ 0     & sin(\phi) & 2sin(\phi) \\ 0 &   0.3818  & -1.0137 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

\begin{bmatrix} d_0     \\ d_1 \\ d_2 \end{bmatrix}

=

\begin{bmatrix} 0\\ 4 \\ 2.1213 \end{bmatrix} $$ $$
 *
 $$ \displaystyle (Eq. 2.3.26)
 * }

For "K" matrix:

0.7071  -0.2130   -0.9372         0    0.7071    1.4142         0    0.3818   -1.0137

For "F" matrix:

0   4.0000    2.1213

(5)

Solving for $$ \underline{d} = \underline{K}^{-1} F\quad $$ at $$ \phi = \frac{\pi}{4} \quad $$ we get

(tried to solve for $$ \phi = \frac{\pi}{2} \quad $$, matrix approx. value came if not exact, then really close to exact and the warning of matlab was issued about singularity)

d = 1.7193   5.6137    0.0216

(6)

Constructed $$ u^h \quad $$ matrix on matlab by using formula $$ u^h = \sum_{i=0}^{n=2} d_ib_i $$

0.02156*cos(phi + 2.0*x) + 5.614*cos(phi + x) + 1.719*cos(phi)

Exact solution:

Given a general expression for 2nd Order ODE

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

$$ a_2(x)y'' + a_1(x)y + a_0(x)y = g(x) \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.27)
 * }

and taking a more specific case ("Self adjoin") for this specific problem, we have

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

$$ [a_2(x)y']' + a_0(x)y = g(x) \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.17)
 * }

$$ g(x) = 0 \Rightarrow \quad $$ no forcing function (steady state)

$$ a_0(x)y = f(x) = 3 \quad $$ (given)

$$ a_2 = 2 \quad $$ (given)

Therefore $$ Eq. 2.3.17 $$ becomes

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

 * style="width:95%" |
 * style="width:95%" |

$$ [2y']' + 3 = 0 \quad $$ $$
 *
 $$ \displaystyle (Eq. 2.3.28)
 * }

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

 * style="width:95%" |
 * style="width:95%" |

$$ \frac{du}{dx}(\frac{du}{dx}2) + 3 = 0 $$ $$
 *
 $$ \displaystyle
 * }

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

$$ \int \frac{du}{dx}2 = \int -3 dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ \frac{du}{dx}2 = -3x + c_1' \Rightarrow \frac{du}{dx} = -1.5x + c_1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.29)
 * }

at B.C of $$ -\frac{du}{dx}(0) = 4 \Rightarrow $$

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

 * style="width:95%" |
 * style="width:95%" |

$$ -4 = -1.5(0) + c_1 \Rightarrow c_1 = -4$$ $$
 *
 $$ \displaystyle
 * }

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

 * style="width:95%" |
 * style="width:95%" |

$$ \int {du} = \int (-1.5x - 4) dx $$ $$
 *
 $$ \displaystyle
 * }

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

$$ u = \frac{-1.5}{2}x^2 - 4x + c_2 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.30)
 * }

at B.C of $$ u(1) = 0 \Rightarrow $$

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

 * style="width:95%" |
 * style="width:95%" |

$$ 0 = \frac{-1.5}{2}(1)^2 - 4(1) + c_2 \Rightarrow c_2 = -4.75$$ $$
 *
 $$ \displaystyle
 * }

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

$$ u(x) = -0.75x^2 - 4x - 4.75 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 2.3.31)
 * }

So taken in account the exact solution formula, different plot were generated for different n values (n = 2, 4, 6)

Different 'u' values plotted on 1 graph for different member number 'n'

Highly zoomed in view, by zooming in it is clear that n =4 is good enough approximation for these series.

Error came out to be ex = n=2       n=4       n=6 -0.2291  -0.0102    0.0000

--Eml5526.s11.team5.JA 20:30, 2 February 2011 (UTC)

= References =

= Contributors =

--Eml5526.s11.team5.JA 20:29, 2 February 2011 (UTC)

--Eml5526.s11.team5.savery 20:33, 2 February 2011 (UTC)

--Eml5526.s11.team5.mcdaniel 20:35, 2 February 2011 (UTC)

--Eml5526.s11.team5.srv 20:36, 2 February 2011 (UTC)

--Lokeshdahiya 20:41, 2 February 2011 (UTC)

--Eml5526.s11.team5.vijay 20:42, 2 February 2011 (UTC)/>

--Eml5526.s11.team5.smith 20:50, 2 February 2011 (UTC)

title: User:Eml5526.s11.team5/sub2


text:
Talk:Solution Architecture

Solution Architecture should be a part of the school of business
It focuses more on the development of business solutions than technology solutions. Much of the curriculum will be in developing the business and architectural skills and knowledge to analyse, design and implement business solutions. The curriculum is focused on the middle ground between the business and leadership and the technologists doing the implementation. - Peterrawsthorne 18:21, 25 January 2011 (UTC)

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

title: Talk:Solution Architecture


title: File:Fe1.s11.tm7.HW1.fig2.v1.png


title: File:Fe1.s11.tm7.HW1.fig2.v2.png


text:
User talk:Eml5526.s11.team7.flater

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title: User talk:Eml5526.s11.team7.flater


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User:2crudedudes

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