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Thread: Fourier Series - Wave Eqn

  1. #1
    Rhymes with Orange Chris L T521's Avatar
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    Fourier Series - Wave Eqn

    I just have a quick question:

    I'm evaluating a Fourier coefficient for the following series:

    $\displaystyle y(x,t)=\sum_{n=1}^{\infty}B_n\sin\left(\frac{n\pi a}{L}t\right)\sin\left(\frac{n\pi}{L}x\right)$

    Where $\displaystyle B_n=\frac{2}{n\pi a}\int_0^L \psi(x)\sin\left(\frac{n\pi}{L}x\right)\,dx$ and $\displaystyle \psi \left( x \right) = \left\{ \begin{array}{cl}
    0 &0 \leqslant x \leqslant \frac{{L - a}}
    {2} \\
    1 & \frac{{L - a}}
    {2} < x < \frac{{L + a}}
    {2} \\
    0 & \,\,\frac{{L + a}}
    {2} < x < L \\
    \end{array} \right.
    $

    I'm going to show my work:

    I said that

    $\displaystyle \begin{aligned}
    B_n&=\frac{2}{n\pi a} \left[\int_0^{\frac{L-a}{2}}(0)\sin\left(\frac{n\pi}{L}x\right)\,dx+\int _{\frac{L-a}{2}}^{\frac{L+a}{2}}\sin\left(\frac{n\pi}{L}x\ri ght)\,dx+ \int_{\frac{L+a}{2}}^L(0)\sin\left(\frac{n\pi}{L}x \right)\,dx\right] \\
    &=\frac{2}{n\pi a}\int_{\frac{L-a}{2}}^{\frac{L+a}{2}}\sin\left(\frac{n\pi}{L}x\ri ght)\,dx
    \end{aligned}$

    $\displaystyle \begin{aligned}
    B_n&=
    \frac{2}{n\pi a}\left.\left[-\frac{L}{n\pi}\cos\left(\frac{n\pi}{L}x\right)\rig ht]\right|_{\frac{L-a}{2}}^{\frac{L+a}{2}} \\
    &=\frac{-2L}{n^2\pi^2a}\left[\cos\left(\frac{n\pi}{L}\cdot\frac{L+a}{2}\right)-\cos\left(\frac{n\pi}{L}\cdot\frac{L-a}{2}\right)\right]
    \end{aligned}
    $

    This is what I did to simplify what was in the brackets:

    $\displaystyle \cos\left(\frac{n\pi}{2}+\frac{n\pi a}{2L}\right)-\cos\left(\frac{n\pi}{2}-\frac{n\pi a}{2L}\right)$

    $\displaystyle \implies \cos\frac{n\pi}{2}\cos\frac{n\pi a}{2L}-\sin\frac{n\pi}{2}\sin\frac{n\pi a}{2L}-\cos\frac{n\pi}{2}\cos\frac{n\pi a}{2L}-\sin\frac{n\pi}{2}\sin\frac{n\pi a}{2L}$

    $\displaystyle \implies -2\sin\frac{n\pi}{2}\sin\frac{n\pi a}{2L}$

    Multiplying by $\displaystyle -\frac{2L}{n^2\pi^2a}$, I get:

    $\displaystyle \frac{4L}{n^2\pi^2a}\sin\frac{n\pi}{2}\sin\frac{n\ pi a}{2L}$

    Now this is where I'm somewhat in the dark. Since I'm plugging in values of n, where $\displaystyle n\in \mathbb{N}$, into $\displaystyle \sin\frac{n\pi}{2}$, I notice when n is odd, I get either 1 or -1, and when n is even, I get zero. Thus, I rewrote my answer as:

    $\displaystyle \frac{4L}{n^2\pi^2a}\sin\frac{(2n-1)\pi}{2}\sin\frac{n\pi a}{2L}$

    I noticed that $\displaystyle \sin\frac{(2n-1)\pi}{2}$ could be written as $\displaystyle (-1)^{n+1}; \text{ } n\in\mathbb{N}$

    So then I saw that $\displaystyle B_n=\frac{4(-1)^{n+1}L}{n^2\pi^2a}\sin\frac{n\pi a}{2L}$

    Thus, my series solution would be

    $\displaystyle y(x,t)=\frac{4L}{\pi^2a}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}\sin\left(\frac{n\pi a}{2L}\right)\sin\left(\frac{n\pi a}{L}t\right)\sin\left(\frac{n\pi}{L}x\right)$

    Is this correct??

    However, my question is this: when I rewrote $\displaystyle \sin\frac{n\pi}{2}$ as $\displaystyle \sin\frac{(2n-1)\pi}{2}$, do I need to replace the other n's in my equation with 2n-1? meaning, would my answer for $\displaystyle B_n$ be $\displaystyle \frac{4L}{(2n-1)^2\pi^2a}\sin\frac{(2n-1)\pi}{2}\sin\frac{(2n-1)\pi a}{2L}$????

    I'd appreciate any help.
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  2. #2
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    However, my question is this: when I rewrote $\displaystyle \sin\frac{n\pi}{2}$ as $\displaystyle \sin\frac{(2n-1)\pi}{2}$, do I need to replace the other n's in my equation with 2n-1? meaning, would my answer for $\displaystyle B_n$ be $\displaystyle \frac{4L}{(2n-1)^2\pi^2a}\sin\frac{(2n-1)\pi}{2}\sin\frac{(2n-1)\pi a}{2L}$????
    Yes. Think about it: with

    $\displaystyle B_n = \frac{4L}{n^2\pi^2a}\sin\frac{n\pi}{2}\sin\frac{n\ pi a}{2L}$

    any time $\displaystyle n$ is even, this term evaluates to zero, as you observed. The only remaining terms will be those for which $\displaystyle n$ is odd. The values of $\displaystyle \frac{4L}{n^2\pi^2a}\text{ and }\sin\frac{n\pi a}{2L}$ do not matter for even $\displaystyle n$, because they're being multiplied by zero. However, with

    $\displaystyle B_n = \frac{4L}{n^2\pi^2a}\sin\frac{(2n-1)\pi}{2}\sin\frac{n\pi a}{2L}$

    this changes. Now, the values of $\displaystyle \frac{4L}{n^2\pi^2a}\text{ and }\sin\frac{n\pi a}{2L}$ will affect the result regardless of whether $\displaystyle n$ is even or odd.

    The rest of your work looks good to me, but I admit that I only gave a quick look through it.
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