User:Egm4313.s12.team11.arrieta/Report 7

Problem R7.2
Solved by Francisco Arrieta

Problem Statement
Plot the truncated series $$ u(x,t)=\sum_{j=1}^{n}a_j\cos C\omega_j t \sin \omega_j x \!$$ with $$ n=5 \!$$ and for:

$$ t=\alpha P_1=\alpha \frac{2\pi}{C\omega_1}=\alpha \frac{2L}{C} \!$$

$$ \alpha=0.5, 1, 1.5, 2 \!$$

Solution
Using:

$$ \left\{\begin{matrix} f(x)=x(x-2) \\ g(x)=0 \\ C=3 \\ L=4

\end{matrix}\right. \!$$

Then:

$$ a_j=\frac{2}{L}\int_{0}^{L}f(x)\sin \omega_j x dx \!$$


 * $$ =2\left [ \frac{(-1)^j-1}{\pi^3j^3} \right ] \!$$

$$ \therefore a_j=0 \!$$ for all even values of j

Plugging back to the truncated series:

$$ u(x,t)=\sum_{j=1}^{n}2\left [ \frac{(-1)^j-1}{\pi^3j^3} \right ]\cos [C \frac{j\pi}{L} \alpha \frac{2L}{C}] \sin[\frac{j\pi}{L}x] \!$$


 * $$ =\sum_{j=1}^{n}2\left [ \frac{(-1)^j-1}{\pi^3j^3} \right ]\cos (\alpha j2\pi) \sin(\frac{j\pi}{2}x) \!$$

For $$ n=5 \!$$ :

$$ u(x,t)=[\frac{-4}{\pi^3}\cos (2\pi \alpha) \sin(\frac{\pi x}{2})]+[\frac{-4}{27\pi^3}\cos (6\pi \alpha) \sin(\frac{3\pi x}{2})]+[\frac{-4}{125\pi^3}\cos (10\pi \alpha) \sin(\frac{5\pi x}{2})] \!$$

When $$ \alpha=0.5 \!$$ :



When $$ \alpha=1 \!$$ :



When $$ \alpha=1.5 \!$$ :



When $$ \alpha=2 \!$$ :