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Mathematics LibreTexts

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.

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