Find the Fourier Series Solution for the case

Question

$u(x,0) = \frac{\pi}{2} - |\frac{\pi}{2} - x | $ , $u_t(x,0)$ for $0 \leq x \leq pi$.

Hint: Vibration of String Plucked at Center

and calculate its energy.

Please help lol

Answer 1

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Energy $\ds{\varepsilon}$ is conserved. So, if $\ds{\rho}$ is the string linear density and $\ds{T}$ is the string tension:

\begin{align} \varepsilon & = \int_{0}^{\pi}{1 \over 2}\,\rho\,\mrm{u}_{t}^{2}\pars{x,0}\,\dd x \\[5mm] & + \int_{0}^{\pi/2}{1 \over 2}\,T \braces{\totald{\bracks{\pi/2 - \pars{\pi/2 - x}}}{x}}^{2}\,\dd x + \int_{\pi/2}^{\pi}{1 \over 2}\,T \braces{\totald{\bracks{\pi/2 + \pars{\pi/2 - x}}}{x}}^{2}\,\dd x \\[5mm] & = \bbx{\ds{% {1 \over2}\,\rho\int_{0}^{\pi}\mrm{u}_{t}^{2}\pars{x,0}\,\dd x + {1 \over 2}\,\pi T}} \end{align}

Answer 2

IMPORTANT NOTE: This post is only long because it contains a detailed step by step solution on finding TWO different Fourier series for the same function: a cosine series and a sine series.

In order to find a Fourier series for the above function, we split it up into two parts:

$$ f(x)= \begin{cases} x &\text{if }0\leq x\leq \frac{\pi}{2}\\ \pi-x &\text{if } \frac{\pi}{2} \leq x \leq \pi \end{cases} $$

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In order to find the sine series, we compute the following integral: $$\frac{2}{\pi}\int_0^\pi f(x)\sin(nx)\,dx=\frac{2}{\pi}\left[\int_0^\frac{\pi}{2} x\sin(nx)\,dx +\int_\frac{\pi}{2}^\pi \left(\pi-x\right)\sin(nx)\,dx\right]$$ For the right integral, we make the substitution $u=\pi-x$:

\begin{align} \frac{2}{\pi}\left[\int_0^\frac{\pi}{2} x\sin(nx)\,dx -\int_\frac{\pi}{2}^0 u\sin\left(n\left(\pi-u\right)\right)\,du\right]\\=\frac{2}{\pi}\left[\int_0^\frac{\pi}{2} x\sin(nx)\,dx +\int_0^\frac{\pi}{2} u\sin\left(n\left(\pi-u\right)\right)\,du\right] \end{align} Now we can use the fact that $\sin\left(n\pi-nu\right)=\left(-1\right)^{n+1}\sin(nu)$: $$\frac{2}{\pi}\left[\int_0^\frac{\pi}{2} x\sin(nx)\,dx +\left(-1\right)^{n+1}\int_0^\frac{\pi}{2} u\sin\left(nu\right)\,du\right]$$ Combining both integrals: $$\frac{2}{\pi}\int_0^\frac{\pi}{2}x\sin(nx)\left(1+\left(-1\right)^{n+1}\right)\,dx$$ When $n$ is even, the integral is $0$. When $n$ is odd, $\left(1+\left(-1\right)^{n+1}\right)=2$: $$\frac{4}{\pi}\int_0^\frac{\pi}{2}x\sin(nx)\,dx$$ Integrating by parts: $$\frac{4}{\pi}\left[\left[\frac{-x\cos(nx)}{n}\right]_0^\frac{\pi}{2}+\int_0^\frac{\pi}{2}\frac{\cos(nx)}{n}\,dx\right]$$ Using the fact that $\cos(n\pi/2)$ is zero when $n$ is odd: $$\frac{4}{\pi}\int_0^\frac{\pi}{2}\frac{\cos(nx)}{n}\,dx=\frac{4}{\pi}\left[\frac{\sin(nx)}{n^2}\right]^\frac{\pi}{2}_0$$ Using the fact that $\sin\left(n\pi/2\right)$ is $1$ if $n\equiv 1 \pmod 4$ and $-1$ if $n\equiv 3\pmod 4$: $$\frac{4}{\pi n^2}\cdot\left(-1\right)^{(n-1)/2}$$ From this, we obtain the following Fourier series: $$\frac{4}{\pi}\sum_{n=0}^{\infty}(-1)^{n}\frac{\sin\left((2n+1)x\right)}{\left(2n+1\right)^2}$$

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In order to find the cosine series, we compute the following integral: $$\frac{2}{\pi}\int_0^\pi f(x)\cos(nx)\,dx=\frac{2}{\pi}\left[\int_0^\frac{\pi}{2} x\cos(nx)\,dx +\int_\frac{\pi}{2}^\pi \left(\pi-x\right)\cos(nx)\,dx\right]$$ For the right integral, we make the substitution $u=\pi-x$: \begin{align} \frac{2}{\pi}\left[\int_0^\frac{\pi}{2} x\cos(nx)\,dx -\int_\frac{\pi}{2}^0 u\cos\left(n\left(\pi-u\right)\right)\,du\right]\\=\frac{2}{\pi}\left[\int_0^\frac{\pi}{2} x\cos(nx)\,dx +\int_0^\frac{\pi}{2} u\cos\left(n\left(\pi-u\right)\right)\,du\right] \end{align} Using the fact that $\cos(n\pi-nu)=(-1)^n\cos(nu)$: $$\frac{2}{\pi}\left[\int_0^\frac{\pi}{2} x\cos(nx)\,dx +(-1)^n\int_0^\frac{\pi}{2} u\cos\left(nu\right)\,du\right]$$ Combining both integrals: $$\frac{2}{\pi}\int_0^\frac{\pi}{2}x\cos(nx)\left(1+\left(-1\right)^n\right)\,dx$$ When $n$ is odd, the integral is $0$. When $n$ is even, $\left(1+\left(-1\right)^{n}\right)=2$: $$\frac{4}{\pi}\int_0^\frac{\pi}{2}x\cos(nx)\,dx$$ Integrating by parts: $$\frac{4}{\pi}\left[\left[\frac{x\sin(nx)}{n}\right]_0^\frac{\pi}{2}-\int_0^\frac{\pi}{2}\frac{\sin(nx)}{n}\,dx\right]$$ Using the fact that $\sin(n\pi/2)$ is zero when $n$ is even: $$-\frac{4}{\pi}\int_0^\frac{\pi}{2}\frac{\sin(nx)}{n}\,dx=\frac{4}{\pi}\left[\frac{\cos(nx)}{n^2}\right]_0^\frac{\pi}{2}$$ Using the fact that $\cos\left(n\pi/2\right)$ is $1$ if $n\equiv 0 \pmod 4$ and $-1$ if $n\equiv 2\pmod 4$: $$\frac{4}{\pi n^2}\cdot\left(\left(-1\right)^{n/2}-1\right)$$ This is equal to zero whenever $n\equiv 0 \pmod 4$. Otherwise, when $n\equiv 2 \pmod 4$, the above expression is equal to the following: $$-\frac{8}{\pi n^2}$$ Now, all we have to do is compute the following integral: $$\frac{2}{\pi}\int_0^\pi f(x)\,dx$$ Using the area formula for a triangle, we find that this integral is equal to $\frac{\pi}{2}$. From this, we obtain the following Fourier series: $$\frac{\pi}{4}-\frac{8}{\pi}\sum_{n=0}^{\infty}\frac{\cos((4n+2)x)}{(4n+2)^2}$$