4.5.1: Oscillation of a String

Let $u(x,t)$, $x\in[a,b]$, $t\in\mathbb{R}^1$, be the deflection of a string, see Figure 1.1.1 from Chapter 1. Assume the deflection occurs in the $(x,u)$-plane. This problem is governed by the initial-boundary value problem
$$\begin{eqnarray} \label{string1}\tag{4.5.1.1} u_{tt}(x,t)&=&u_{xx}(x,t)\ \ \mbox{on}\ (0,l)\\ \label{string2} \tag{4.5.1.2} u(x,0)&=&f(x)\\ \label{string3} \tag{4.5.1.3} u_t(x,0)&=&g(x)\\ \label{string4} \tag{4.5.1.4} u(0,t)&=&u(l,t)=0. \end{eqnarray}$$
Assume the initial data $f$, $g$ are sufficiently regular. This implies compatibility conditions $f(0)=f(l)=0$ and $g(0)=g(l)$.

Fourier's method

To find solutions of differential equation ($\ref{string1}$) we make the separation of variables ansatz
$$u(x,t)=v(x)w(t).$$
Inserting the ansatz into ($\ref{string1}$) we obtain
$$v(x)w''(t)=v''(x)w(t),$$
or, if $v(x)w(t)\not=0$,
$$\frac{w''(t)}{w(t)}=\frac{v''(x)}{v(x)}.$$
It follows, provided $v(x)w(t)$ is a solution of differential equation ($\ref{string1}$) and $v(x)w(t)\not=0$,
$$\frac{w''(t)}{w(t)}=const.=:-\lambda$$
and
$$\frac{v''(x)}{v(x)}=-\lambda$$
since $x,\ t$ are independent variables.

Assume $v(0)=v(l)=0$, then $v(x)w(t)$ satisfies the boundary condition ($\ref{string4}$). Thus we look for solutions of the eigenvalue problem
$$\begin{eqnarray} \label{ewastring1}\tag{4.5.1.5} -v''(x)&=&\lambda v(x)\ \ \mbox{in}\ (0,l)\\ \label{ewastring2} \tag{4.5.1.6} v(0)&=&v(l)=0, \end{eqnarray}$$
which has the eigenvalues
$$\lambda_n=\left(\frac{\pi}{l}n\right)^2, \ \ n=1,2,\ldots,$$
and associated eigenfunctions are
$$v_n=\sin\left(\frac{\pi}{l}nx\right).$$
Solutions of
$$-w''(t)=\lambda_n w(t)$$
are
$$\sin(\sqrt{\lambda_n}t),\ \ \ \cos(\sqrt{\lambda_n}t).$$
Set
$$w_n(t)=\alpha_n\cos(\sqrt{\lambda_n}t)+\beta_n\sin(\sqrt{\lambda_n}t),$$
where $\alpha_n,\ \beta_n\in\mathbb{R}^1$.
It is easily seen that $w_n(t)v_n(x)$ is a solution of differential equation ($\ref{string1}$), and, since ($\ref{string1}$) is linear and homogeneous, also (principle of superposition)
$$u_N=\sum_{n=1}^Nw_n(t)v_n(x)$$
which satisfies the differential equation ($\ref{string1}$) and the boundary conditions ($\ref{string4}$). Consider the formal solution of ($\ref{string1}$), ($\ref{string4}$)
$$\begin{equation} \label{string5} \tag{4.5.1.7} u(x,t)=\sum_{n=1}^\infty \left(\alpha_n\cos(\sqrt{\lambda_n}t)+\beta_n\sin(\sqrt{\lambda_n}t)\right)\sin\left(\sqrt{\lambda_n}x\right). \end{equation}$$
''Formal'' means that we know here neither that the right hand side converges nor that it is a solution of the initial-boundary value problem. Formally, the unknown coefficients can be calculated from initial conditions ($\ref{string2}$), ($\ref{string3}$) as follows. We have
$$u(x,0)=\sum_{n=1}^\infty\alpha_n\sin(\sqrt{\lambda_n}x)=f(x).$$
Multiplying this equation by $\sin(\sqrt{\lambda_k}x)$ and integrate over $(0,l)$, we get
$$\alpha_n\int_0^l\ \sin^2(\sqrt{\lambda_k}x)\ dx=\int_0^l\ f(x)\sin(\sqrt{\lambda_k}x)\ dx.$$
We recall that
$$\int_0^l\ \sin(\sqrt{\lambda_n}x)\sin(\sqrt{\lambda_k}x)\ dx=\frac{l}{2}\delta_{nk}.$$
Then
$$\begin{equation} \label{string6} \tag{4.5.1.8} \alpha_k=\frac{2}{l}\int_0^l\ f(x)\sin\left(\frac{\pi k}{l} x\right)\ dx. \end{equation}$$
By the same argument it follows from
$$u_t(x,0)=\sum_{n=1}^\infty\beta_n\sqrt{\lambda_n}\sin(\sqrt{\lambda_n}x)=g(x)$$
that
$$\begin{equation} \label{string7} \tag{4.5.1.9} \beta_k=\frac{2}{k\pi}\int_0^l\ g(x)\sin\left(\frac{\pi k}{l} x\right)\ dx. \end{equation}$$

Under additional assumptions $f\in C_0^4(0,l)$, $g\in C_0^3(0,l)$ it follows that the right hand side of ($\ref{string5}$), where $\alpha_n$, $\beta_n$ are given by ($\ref{string6}$) and ($\ref{string7}$), respectively, defines a classical solution of ($\ref{string1}$)-($\ref{string4}$) since under these assumptions the series for $u$ and the formal differentiate series for $u_t$, $u_{tt}$, $u_x$, $u_{xx}$ converges uniformly on $0\le x\le l$, $0\le t\le T$, $0<T<\infty$ fixed, see an exercise.

Contributors and Attributions

• Prof. Dr. Erich Miersemann (Universität Leipzig)

• Integrated by Justin Marshall.