I was given a guitar string of length 1 with fixed endpoints. My $f(x)$ is $2x$ if $(x \le 0.5)$ and $-2x+2$ if $(x \gt 0.5)$. My initial velocity is 0. $f(x)$ is the initial position
I was first asked to find first 3 coefficients in sine Fourier series. By doing $a_k = \frac{2}{l} \int_0^1 f(x)sin(\frac{\pi kx}{l})dx$, I got that $a_1 = \frac{8}{\pi^2}$, $a_2 = 0$, and $a_3 = -\frac{8}{9\pi^2}$.
Now, the question asks for solution of this initial boundary problem. Basically, I have $u_t(x,0) = 0$ and $u(0,t) = u(1,t) = 0$. I learned that the general solution looks like the following
$u(x,t) = \sum_{k=1}^{\infty} sin(\frac{\pi kx}{l})(a_k sin(\frac{\pi kx}{l}) + b_k cos(\frac{\pi kx}{l}))$.
Now, I know $l=1$ and I can simply substitute 1,2, and 3 to $k$. I already have corresponding $a_k$ to this. However, I am not sure if there is a $b_k$. I think that all $b_k = 0$ leading me to conclude that $u$ only has $sin$ in it. However, I am not sure how to arrive at the solution. Could someone help? Thank you.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align} \mrm{u}\pars{x,t} & = \sum_{n = 1}^{\infty}\mrm{a}_{n}\pars{t}\sin\pars{n\pi x} \\[2mm] \mbox{such that} & \sum_{n = 1}^{\infty} \braces{\mrm{a}_{n}\pars{t}\bracks{-n^{2}\pi^{2}\sin\pars{n\pi x}} - \ddot{\mrm{a}}_{n}\pars{t}\sin\pars{n\pi x}} \\[2mm] \implies 0 & = \bracks{\totald[2]{}{t} + \pars{n\pi}^{2}}\mrm{a}_{n}\pars{t} \end{align}
$$ \implies \left\{\begin{array}{rcl} \ds{\mrm{u}\pars{x,t}} & \ds{=} & \ds{\sum_{n = 1}^{\infty}\bracks{% \mrm{a}_{n}\pars{0}\cos\pars{n\pi t} + \dot{\mrm{a}}_{n}\pars{0} {\sin\pars{n\pi t} \over n\pi}}\sin\pars{n\pi x}} \\ \ds{\mrm{u}_{t}\pars{x,0}} & \ds{=} & \ds{\sum_{n = 1}^{\infty} \dot{\mrm{a}}_{n}\pars{0}\sin\pars{n\pi x} \implies \dot{\mrm{a}}_{n}\pars{0} = 0} \\ \ds{\mrm{u}\pars{x,0}} & \ds{=} & \ds{\sum_{n = 1}^{\infty} \mrm{a}_{n}\pars{0}\sin\pars{n\pi x} \implies \mrm{a}_{n}\pars{0} = 2\int_{0}^{1}\mrm{u}\pars{x,0} \sin\pars{n\pi x}\dd x} \end{array}\right. $$
Then, \begin{align} \mrm{a}_{n}\pars{0} & = 2\int_{0}^{1/2}2x\sin\pars{n\pi x}\,\dd x + 2\int_{1/2}^{1}\pars{2 - 2x}\sin\pars{n\pi x}\,\dd x \\[5mm] & = {8\sin\pars{n\pi/2} \over n^{2}\pi^{2}} = \left\{\begin{array}{lcl} \ds{0} & \mbox{if} & \ds{n}\ \mbox{is}\ even \\ \ds{8\pars{-1}^{\pars{n - 1}/2} \over n^{2}\pi^{2}} & \mbox{if} & \ds{n}\ \mbox{is}\ odd \end{array}\right. \end{align}
$$ \bbx{\mrm{u}\pars{x,t} = {8 \over \pi^{2}}\sum_{n = 0}^{\infty} {\pars{-1}^{n} \over \pars{2n + 1}^{2}}\, \sin\pars{\bracks{2n + 1}\pi x} \cos\pars{\bracks{2n + 1}\pi t}} $$