
# 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})

\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).

''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

\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.

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

\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.

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

• Integrated by Justin Marshall.