5.2: Application of eigenfunction series

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Contributed by Jiří Lebl
Assistant Professor (Mathematics) at Oklahoma State University

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5.2 Application of eigenfunction series

The eigenfunction series can arise even from higher order equations. Suppose we have an elastic beam (say made of steel). We will study the transversal vibrations of the beam. That is, assume the beam lies along the $x$ axis and let $y(x,t)$ measure the displacement of the point $x$ on the beam at time $t$. See Figure 5.2.

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Figure 5.2: Transversal vibrations of a beam.

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

Jiří Lebl (Oklahoma State University).These pages were supported by NSF grants DMS-0900885 and DMS-1362337.