# Vibrating String with Fixed Ends

Until now, we have been working only with the heat equation. Today, we are going to solve the one-dimensional wave equation with homogeneous boundary conditions and no sources, namely,
$$\begin{align}
u_{tt}&=c^2u_{xx},\\
u(0,t)&=0,\quad u(L,t)=0,\\
u(x,0)&=f(x),\quad u_t(x,0)=g(x).
\end{align}$$
As before, we want to use the method of separation of variables by setting $u(x,t)=\phi(x)h(t)$ and substituting above:
$$
\phi h''=c^2\phi''h.
$$
We want to ignore trivial solutions. This implies our boundary conditions are $\phi(0)=0$, and $\phi(L)=0$.

Separating variables yields
$$
\frac{\phi''}{\phi}=\frac{h''}{c^2h}\color{blue}{=-\lambda},
$$
Because functions of distinct independent variables can **only** be equal if they equate to the same constant, we have introduced the equality in blue (the negative sign is purely out of convenience later on).

We now have a system of two ordinary differential equations: $$\begin{align} \phi''&=-\lambda\phi,\tag{1}\\ h''&=-\lambda c^2h.\tag{2} \end{align}$$ Equation $(1)$ has three cases:

- $\lambda<0$: $$ \phi(x)=c_1\sinh\sqrt{-\lambda}x+c_2\cosh\sqrt{-\lambda}x. $$ Applying the boundary conditions to this yields $$\begin{align} \phi(0)&=c_2=0,\\ \phi(L)&=c_1\sinh\sqrt{-\lambda}L=0\Longrightarrow c_1=0. \end{align}$$ This is the trivial solution, so we drop it.
- $\lambda=0$: $$ \phi(x)=c_1x+c_2. $$ Applying the boundary conditions to this yields $$\begin{align} \phi(0)&=c_2=0,\\ \phi(L)&=c_1L=0\Longrightarrow c_1=0. \end{align}$$ This is the trivial solution, so we drop it.
- $\lambda>0$: $$ \phi(x)=c_1\sin\sqrt{\lambda}x+c_2\cos\sqrt{\lambda}x. $$ Applying the boundary conditions to this yields $$\begin{align} \phi(0)&=c_2=0,\\ \phi(L)&=c_1\sin\sqrt{\lambda}L=0\\ &\Longrightarrow\lambda_n=\left(\frac{n\pi}{L}\right)^2,\\ \phi_n(x)&=C_n\sin\frac{n\pi x}{L}. \end{align}$$

**principle of superposition**, we find that the solution to this partial differential equation is $$ \color{blue}{u(x,t)=\sum_{n=1}^{\infty}\left(G_n\sin \frac{n\pi ct}{L}+F_n\cos \frac{n\pi ct}{L}\right)\sin\frac{n\pi x}{L}.} $$ The coefficients $F_n$ and $G_n$ can be found by applying the initial conditions: $$ u(x,0)=\sum_{n=1}^{\infty}F_n\sin\frac{n\pi x}{L}=f(x). $$ $$\begin{align} u_t(x,t)&=\sum_{n=1}^{\infty}\frac{n\pi c}{L}\left(G_n\cos \frac{n\pi ct}{L}-F_n\sin \frac{n\pi ct}{L}\right)\sin\frac{n\pi x}{L},\\ u_t(x,0)&=\sum_{n=1}^{\infty}G_n\frac{n\pi c}{L}\sin\frac{n\pi x}{L}=g(x). \end{align}$$ Finally, applying**orthogonality principles**to the above equations to find $F_n$ and $G_n$ yields $$\begin{align} \color{blue}{F_n}&\color{blue}{=\frac{2}{L}\int_{0}^{L}f(x)\sin\frac{n\pi x}{L}\,dx,}\\ \color{blue}{G_n}&\color{blue}{=\frac{2}{n\pi c}\int_{0}^{L}g(x)\sin\frac{n\pi x}{L}\,dx.} \end{align}$$ Letting $c=1$, $L=1$, $f(x)=\sin x$ and $g(x)=0$ gives us the following wave (the approximation is very low):

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