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# 5.2: Application of eigenfunction series

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## 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

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