User:EML4500.f08.JAMAMA/FE/savery

Problem

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Consider $$ F=\left \{ 1,x,x^2,x^3,x^4 \right \} $$ on the interval $$ \Omega =[0,1] $$ $$
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 *  $$ \displaystyle
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Find

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1. Construct the $$ \Gamma(F)$$ matrix and observe its propeties.
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 * 2. Find the determinant of $$ \Gamma(F)$$.
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3. Conclude if F is an orthogonal family., i.e., $$ \Gamma_{ij} = \delta_{ij} $$
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1)

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The Gram Matrix is defined as:
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 * $$\Gamma_{ij}==\int_\Omega b_i(x)b_j(x)\,dx$$

$$
 *  $$ \displaystyle (eq. 1.1)
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For the case where $$ F=\left \{ 1,x,x^2,x^3,x^4 \right \} $$ $$ 
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 *  $$ \displaystyle
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$$ b_1(x)=1$$ $$
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 *  $$ \displaystyle
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$$ b_2(x)=x $$ $$
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 *  $$ \displaystyle
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$$  b_3(x)=x^2  $$ $$
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 *  $$ \displaystyle
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$$  b_4(x)=x^3  $$ $$
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 * <p style="text-align:right"> $$ \displaystyle
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$$ b_5(x)=x^4  $$ $$
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 * <p style="text-align:right"> $$ \displaystyle
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Now we need to calculate each element in the $$ \Gamma $$ Matrix using eqn. 1.1: $$
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 * <p style="text-align:right"> $$ \displaystyle
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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 $$\Gamma_{ji}$$ $$
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 * <p style="text-align:right"> $$ \displaystyle
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<span id="(1)">
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$$ \Gamma_{1,1} = \int_{0}^{1} 1*1*dx = \left [ x\right ]_{0}^{1} = 1 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle
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<span id="(1)">
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$$ \Gamma_{1,2} = \Gamma_{2,1} = \int_{0}^{1} 1*x*dx = \left [ x^2/2\right ]_{0}^{1} = 1/2 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle
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<span id="(1)">
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$$ \Gamma_{1,3} = \Gamma_{3,1} = \int_{0}^{1} 1*x^2*dx = \left [ x^3/3 \right ]_{0}^{1} = 1/3 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle
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$$ \Gamma_{1,4} = \Gamma_{4,1} = \int_{0}^{1} 1*x^3*dx = \left [ x^4/4 \right ]_{0}^{1} = 1/4 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle
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$$ \Gamma_{1,5} = \Gamma_{5,1} = \int_{0}^{1} 1*x^4*dx = \left [ x^5/5 \right ]_{0}^{1} = 1/5 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle
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<span id="(1)">
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$$ \Gamma_{2,2} = \int_{0}^{1} x*x*dx = \left [ x^3/3\right ]_{0}^{1} = 1/3 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
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$$ \Gamma_{2,3} = \Gamma_{3,2} = \int_{0}^{1} x*x^2*dx = \left [ x^4/4\right ]_{0}^{1} = 1/4 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
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$$ \Gamma_{3,3} = \int_{0}^{1} x^2*x^2*dx = \left [ x^5/5 \right ]_{0}^{1} = 1/5 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle
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<span id="(1)">
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$$ \Gamma_{3,4} = \Gamma_{4,3} = \int_{0}^{1} x^2*x^3*dx = \left [ x^6/6\right ]_{0}^{1} = 1/6 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle
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$$ \Gamma_{3,5} = \Gamma_{5,3} = \int_{0}^{1} x^2*x^4*dx = \left [ x^7/7\right ]_{0}^{1} = 1/7 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle
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$$ \Gamma_{4,4} = \int_{0}^{1} x^3*x^3*dx = \left [ x^7/7\right ]_{0}^{1} = 1/7 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle
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$$ \Gamma_{4,5} = \Gamma_{5,4} = \int_{0}^{1} x^3*x^4*dx = \left [ x^8/8\right ]_{0}^{1} = 1/8 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle
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$$ \Gamma_{5,5} = \Gamma_{5,5} = \int_{0}^{1} x^4*x^4*dx = \left [ x^9/9\right ]_{0}^{1} = 1/9 $$ $$
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 * <p style="text-align:right"> $$ \displaystyle
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Assembling the matrix we get: $$
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\Gamma_ \left (F \right) = \begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5}\\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6}\\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \frac{1}{7}\\ \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \frac{1}{7} & \frac{1}{8}\\ \frac{1}{5} & \frac{1}{6} & \frac{1}{7} & \frac{1}{8} & \frac{1}{9} \end{bmatrix} $$

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The matrix is observed to only be symmetric $$
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2)
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Using the matlab function det(X) the determinant of $$\Gamma$$ can be found to be: $$
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 * $$ \det \left ( \Gamma \right ) = 0 $$

3)
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To conclude if $$\Gamma$$ is orthogonal we compare it to a matrix formed by the Kronecker Delta. $$
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The Kronecker Delta is defined by: $$
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$$\delta_{ij} = \begin{cases} 1, & \mbox{ for i=j} \\ 0, & \mbox{ for i≠j} \end{cases} $$ $$
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Thus the general form of a the matrix is: $$
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$$ \begin{bmatrix} 1    & \cdots & 0      \\ \vdots & \ddots & \vdots \\ 0     & \cdots & 1 \end{bmatrix}
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$$ $$
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Thus the $$ \Gamma $$ Matrix is not orthogonal $$
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--Eml5526.s11.team5.savery 19:49, 2 February 2011 (UTC) = Problem 4.2, Exercise = 3.5, Fish and Belytschko =

Problem
Obtain the Weak Form for the equation of Heat Conduction with the Boundary Condition $$ T(0)= \ 100$$ and $$ \overline{q}= \ hT $$. The condition on the right is a convection condition.

Solution
{| style="width:100%" border="0" Here $$ \quad \tilde{m}(x) =\quad m(x).c$$. This is all due to change in temperature i.e. due to Convection. $$ \mathbf{Boundary \quad Conditions:}$$
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on above Eq. 4.2.7,We get:

{| style="width:100%" border="0" As we know that $$ w \quad $$ has to satisfy Homogeneous Essential Boundary Condition i.e. $$ w(0)= 0 \quad $$. So, Eq.4.2.8 becomes
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