User:Eml5526.sp11.team1.adam/HW3

Solution
The trial solution $$ u\left ( x \right )$$ and the weight function $$ w\left ( x \right )$$ can be written as follows:

$$ u\left ( x \right ) = \alpha_{0} + \alpha_{1}\left ( x-3 \right ) + \alpha_{2}\left ( x-3 \right )^{2}$$

$$ w\left ( x \right ) = \beta_{0} + \beta_{1}\left ( x-3 \right ) + \beta_{2}\left ( x-3 \right )^{2}$$

When:

$$ w \left(3 \right) = 0; \Rightarrow \beta_{0} = 0 $$

and for:

$$ u \left(3 \right) \Rightarrow \alpha_{0} = 0.001 $$

$$ u\left ( x \right )$$ and $$ w\left ( x \right )$$ can therefore be re-written as follows:

$$ u\left ( x \right ) = 0.001 + \alpha_{1}\left ( x-3 \right )+\alpha_{2}\left ( x-3 \right )^{2}$$

$$ w\left( x \right) = \beta_{1} \left(x - 3 \right)+\beta_{2}\left ( x-3 \right )^{2}$$

Whith their respective derivatives:

$$\frac{du}{dx} = \alpha_{1}+2\alpha_{2}\left(x-3\right)$$

$$\frac{dw}{dx} = \beta_{1}+2\beta_{2}\left(x-3\right)$$

Now substituting into the weak form yields:

$$ \int_{1}^{3} AE[\beta_{1}+2\beta_{2}\left(x-3\right)][\alpha_{1}+2\alpha_{2}\left(x-3\right)]dx = -0.1[A(\beta_{1}\left(x-3\right)+\beta_{2}\left(x-3\right)^{2})]_{x=1} + \int_{1}^{3} 2x[\beta_{1}\left(x-3\right)+\beta_{2}\left(x-3\right)^{2}]dx$$

which can be re-arranged as:

$$ \int_{1}^{3} AE[\beta_{1}+2\beta_{2}\left(x-3\right)][\alpha_{1}+2\alpha_{2}\left(x-3\right)]dx - 0.1[A(\beta_{1}\left(x-3\right)+\beta_{2}\left(x-3\right)^{2})]_{x=1} - \int_{1}^{3} 2x[\beta_{1}\left(x-3\right)+\beta_{2}\left(x-3\right)^{2}]dx = 0$$

Evaluating the integral we get:

$$ AE \left(2\alpha_{1}\beta_{1}-4\alpha_{1}\beta_{2}-4\alpha_{2}\beta_{1}+\frac{32\alpha_{2}\beta_{2}}{3}\right)- 0.1[A(4\beta_{2}-2\beta_{1})] - \left(8\beta_{2}-\frac {20\beta_{1}}{3}\right)=0 $$

Factoring out $$\left( \beta_{1}\right)$$ and $$\left( \beta_{2}\right)$$ we get:

$$\beta_{1}\left[2AE\left(\alpha_{1}-2\alpha_{2}\right)-0.2A+\frac{20}{3}\right]+\beta_{2}\left[4AE\left(\frac{8}{3}\alpha_{2}-\alpha_{1}\right)+0.4A-8\right]=0$$

$$\displaystyle {{\beta }_{1}}$$ and $$\displaystyle {{\beta }_{2}}$$ must be zero since they were arbitrarely chosen and because the above equation must hold. Therefore we can write both terms of the above equation equal to zeor as folows:

$$ \left[2AE\left(\alpha_{1}-2\alpha_{2}\right)-0.2A+\frac{20}{3}\right]=0$$

and

$$ \left[4AE\left(\frac{8}{3}\alpha_{2}-\alpha_{1}\right)+0.4A-8\right]=0 $$

Solving for $$\displaystyle {{\alpha }_{1}}$$ and $$\displaystyle {{\alpha }_{2}}$$ we find that:

$$\displaystyle {{\alpha }_{1}}=\frac{\frac{116}{3}+0.2A}{2AE} $$

and

$$\displaystyle {{\alpha }_{2}}=\frac{8}{AE} $$

The solution to the weak form can be written as:

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

$$ u(x)= 0.001+\frac{\frac{116}{3}+0.2A}{2AE}(x-3)-\frac{8}{AE}(x-3)^{2} $$
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 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 10.1) $$
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Checking if equilibrium in the strong form is satisfied
Substituted into the equations:

$$ \frac{d}{dx} \left( AE \frac{du}{dx} \right) + 2x = 0 $$

Yields the following:

$$\displaystyle {2x-16=0}$$

which does not satisfy.

Checking if the natural boundary condition is satisfied
The following equation can be used:

$$ \sigma \left(1 \right) = \left(E \frac{du}{dx} \right)_{x=1} = 0.1 $$