For damped vibrating spring with $\frac{\partial u}{\partial t}(x,0) = g(x) = 0$, show that $u(x,t) = \frac{1}{2} \left[ F(x-ct) + F(x+ct) \right]$

Question

Consider a slightly damped vibrating string that satisfies:

\begin{align*} \rho_0 \frac{\partial^2 u}{\partial t^2} &= T_0 \frac{\partial^2 u}{\partial x^2} - \beta \frac{\partial u}{\partial t} \\ u(0,t) &= 0 \\ u(L,t) &= 0 \\ u(x,0) &= f(x) \\ \frac{\partial u}{\partial t}(x,0) &= g(x) \\ \end{align*}

which leads to the general solution:

\begin{align*} \omega_n &= \sqrt{ \frac{T_0}{\rho_0} \left( \frac{n \pi}{L} \right)^2 - \frac{\beta^2}{4 \rho_0^2}} \\ u(x,t) &\sim e^{-\frac{\beta}{2 \rho_0}t} \sum\limits_{n=1}^\infty (a_n \cos \omega_n t + b_n \sin \omega_n t) \cdot \sin \frac{n \pi x}{L} \\ \end{align*}

If this system is initially at rest, $g(x) = 0$, show that

\begin{align*} u(x,t) &= \frac{1}{2} \left[ F(x-ct) + F(x+ct) \right] \\ \end{align*}

where $F(x)$ is the odd periodic extension of $f(x)$.

Hints:

1. For all $x$, $F(x) = \sum_{n=1}^\infty a_n \sin \frac{n \pi x}{L}$

2. $\sin a \cos b = \frac{1}{2} [\sin (a+b) + \sin (a-b)]$.

My work:

Considering the definition of $f(x)$:

\begin{align*} u(x,t) &\sim e^{-\frac{\beta}{2 \rho_0}t} \sum\limits_{n=1}^\infty (a_n \cos \omega_n t + b_n \sin \omega_n t) \cdot \sin \frac{n \pi x}{L} \\ u(x,0) = f(x) &\sim \sum\limits_{n=1}^\infty a_n \cdot \sin \frac{n \pi x}{L} \\ \end{align*}

which is the definition for $F(x)$. So, that merely gives us the first hint.

Considering the definition of $g(x)$:

\begin{align*} u(x,t) &= e^{-\frac{\beta}{2 \rho_0}t} \sum\limits_{n=1}^\infty (a_n \cos \omega_n t + b_n \sin \omega_n t) \cdot \sin \frac{n \pi x}{L} \\ \frac{\partial u}{\partial t}(x,0) = g(x) = 0 &= -\frac{\beta}{2 \rho_0} \sum\limits_{n=1}^\infty a_n \cdot \sin \frac{n \pi x}{L} + \sum\limits_{n=1}^\infty b_n \omega_n \cdot \sin \frac{n \pi x}{L} \\ 0 &= \sum\limits_{n=1}^\infty \left[b_n \omega_n -\frac{\beta}{2 \rho_0} a_n \right] \cdot \sin \frac{n \pi x}{L} \\ \end{align*}

Which will be true for all $x$ when:

\begin{align*} b_n \omega_n &= \frac{\beta}{2 \rho_0} a_n \\ \end{align*}

Applying the two hints yields:

\begin{align*} F(x) &= \sum_{n=1}^\infty a_n \sin \frac{n \pi x}{L} \\ \frac{1}{2} \left[ F(x-ct) + F(x+ct) \right] &= \frac{1}{2} \sum_{n=1}^\infty a_n \left( \sin \frac{n \pi (x-ct)}{L} + \sin \frac{n \pi (x+ct)}{L} \right) \\ &= \sum_{n=1}^\infty a_n \sin \frac{n \pi x}{L} \cos \frac{n \pi ct}{L} \\ \end{align*}

So, it will be adequate to demonstrate that $u(x,t)$ equals that last expression.

Trying to work backwards, using question marks over the equal signs, to indicate that I haven't proved this but am trying to do so:

\begin{align*} u(x,t) &\stackrel{?}{=} \sum_{n=1}^\infty a_n \sin \frac{n \pi x}{L} \cos \frac{n \pi ct}{L} \\ e^{-\frac{\beta}{2 \rho_0}t} \sum\limits_{n=1}^\infty (a_n \cos \omega_n t + b_n \sin \omega_n t) \cdot \sin \frac{n \pi x}{L} &\stackrel{?}{=} \sum_{n=1}^\infty a_n \sin \frac{n \pi x}{L} \cos \frac{n \pi ct}{L} \\ e^{-\frac{\beta}{2 \rho_0}t} \sum\limits_{n=1}^\infty (a_n \cos \omega_n t + b_n \sin \omega_n t) &\stackrel{?}{=} \sum_{n=1}^\infty a_n \cos \frac{n \pi ct}{L} \\ e^{-\frac{\beta}{2 \rho_0}t} (\cos \omega_n t + \frac{b_n}{a_n} \sin \omega_n t) &\stackrel{?}{=} \cos \frac{n \pi ct}{L} \\ e^{-\frac{\beta}{2 \rho_0}t} (\cos \omega_n t + \frac{\beta}{2 \rho_0 \omega_n} \sin \omega_n t) &\stackrel{?}{=} \cos \frac{n \pi ct}{L} \\ \end{align*}

From here, I'm really stuck on what to try.