Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

5.2: Application of eigenfunction series

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    5.2 Application of eigenfunction series

    The equation that governs this setup is

    \[ a^4 \frac{\partial^4y}{\partial x^4}+\frac{\partial^2y}{\partial t^2}=0,\]

    for some constant \(a>0\).

    Suppose the beam is of length \(1\) simply supported (hinged) at the ends. The beam is displaced by some function \(f(x)\) at time \(t=0\) and then let go (initial velocity is \(0\)). Then \(y\) satisfies:

    \[ a^4y_{xxxx}+y_{tt}=0~~~~(0<x<1,t>0), \\ y(0,t)=y_{xx}(0,t)=0, \\ y(1,t)=y_{xx}(1,t)=0, \\ y(x,0)=f(x),~~~~y_t(x,0)=0. \]

    Again we try \(y(x,t)=X(x)T(t)\) and plug in to get \(a^4X^{(4)}T+XT''=0\) or

    \[\frac{X^{(4)}}{X}=\frac{-T''}{a^4T}=\lambda .\]

    We note that we want \( T''+\lambda a^4T=0\). Let us assume that \(\lambda >0\). We can argue that we expect vibration and not exponential growth nor decay in the \(t\) direction (there is no friction in our model for instance). Similarly \(\lambda =0\) will not occur.

    Exercise \(\PageIndex{1}\):

    Try to justify \(\lambda >0\) just from the equations.

    Write \( \omega^4=\lambda\), so that we do not need to write the fourth root all the time. For \(X\) we get the equation \(X^{(4)}- \omega^4X=0\). The general solution is

    \[ X(x)=Ae^{\omega x}+Be^{- \omega x}+C\sin(\omega x)+D\cos(\omega x).\]

    Now \(0=X(0)A+B+D, 0=X''(0)=\omega^2(A+B-D)\). Hence, \(D=0\) and \(A+B=0\), or \(B=-A\). So we have

    \[ X(x)=Ae^{\omega x}-Ae^{- \omega x}+C\sin(\omega x).\]

    Also \(0=X(1)=A(e^{\omega}-e^{- \omega})+C\sin \omega\), and \(0=X''(1)=A\omega^2(e^{\omega}-e^{- \omega})-C\omega^2\sin \omega\). This means that \(C\sin \omega =0\) and \(A(e^{\omega}-e^{- \omega})=2A\sinh \omega=0\). If \(\omega >0\), then \(\omega \neq 0\) and so \(A=0\). This means that \(C \neq 0\) otherwise \(\lambda\) is not an eigenvalue. Also \(\omega\) must be an integer multiple of \(\pi\). Hence \(\omega =n\pi\) and \(n \geq 1\) (as \(\omega >0\)). We can take \(C=1\). So the eigenvalues are \(\lambda_n=n^4\pi^4\) and the eigenfunctions are \(\sin(n\pi x)\).

    Now \(T''+n^4\pi^4a^4T=0\). The general solution is \( T(t)=A\sin(n^2\pi^2a^2t)+B\cos(n^2\pi^2a^2t)\). But \(T'(0)=0\) and hence we must have \(A=0\) and we can take \(B=1\) to make \(T(0)=1\) for convenience. So our solutions are \( T_n(t)=\cos(n^2\pi^2a^2t)\).

    As the eigenfunctions are just sines again, we can decompose the function \(f(x)\) on \(0<x<1\) using the sine series. We find numbers \(b_n\) such that for \(0<x<1\) we have

    \[ f(x)=\sum_{n=1}^{\infty}b_n\sin(n\pi x).\]

    Then the solution to (5.2.2) is

    \[ y(x,t)=\sum_{n=1}^{\infty}b_nX_n(x)T_n(t)=\sum_{n=1}^{\infty}b_n\sin(n\pi x)\cos(n^2\pi^2a^2t).\]

    The point is that \(X_nT_n\) is a solution that satisfies all the homogeneous conditions (that is, all conditions except the initial position). And since and \(T_n(0)=1\), we have

    \[ y(x,0)=\sum_{n=1}^{\infty}b_nX_n(x)T_n(0)=\sum_{n=1}^{\infty}b_nX_n(x)=\sum_{n=1}^{\infty}b_n\sin(n\pi x)=f(x).\]

    So \(y(x,t)\) solves (5.2.2).

    The natural (circular) frequencies of the system are \(n^2\pi^2a^2\). These frequencies are all integer multiples of the fundamental frequency \(\pi^2a^2\), so we get a nice musical note. The exact frequencies and their amplitude are what we call the timbre of the note.

    The timbre of a beam is different than for a vibrating string where we get “more” of the lower frequencies since we get all integer multiples, \(1,2,3,4,5, \ldots\). For a steel beam we get only the square multiples \(1,4,9,16,25, \ldots\). That is why when you hit a steel beam you hear a very pure sound. The sound of a xylophone or vibraphone is, therefore, very different from a guitar or piano.

    Example \(\PageIndex{1}\):

    Let us assume that \(f(x)= \frac{x(x-1)}{10}\). On \(0<x<1\) we have (you know how to do this by now)

    \[ f(x)= \sum_{ \underset{\rm{n~odd}}{n=1}}^{\infty}\frac{4}{5\pi^3n^3}\sin(n\pi x).\]

    Hence, the solution to (5.2.2) with the given initial position \(f(x)\) is

    \[ y(x,t)= \sum_{ \underset{\rm{n~odd}}{n=1}}^{\infty}\frac{4}{5\pi^3n^3}\sin(n\pi x)\cos(n^2\pi^2a^2t).\]

    Contributors and Attributions