###### Exercise5.2.1

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

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*1 lecture, §10.2 in [EP], exercises in §11.2 in [BD]*

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 \(t\text{.}\) See Figure 6.2.1.

The equation that governs this setup is

\begin{equation*}
a^4 \frac{\partial^4 y}{\partial x^4} + \frac{\partial^2 y}{\partial t^2} = 0,
\end{equation*}

for some constant \(a > 0\text{.}\)

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:

\begin{equation}
\begin{aligned}
& a^4 y_{xxxx} + y_{tt} = 0 \quad (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), \quad y_{t}(x,0) = 0 .
\end{aligned}\label{appeig_beameq}\tag{5}
\end{equation}

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

\begin{equation*}
\frac{X^{(4)}}{X} = \frac{- T''}{a^4T} = \lambda .
\end{equation*}

We note that we want \(T'' + \lambda a^4T = 0\text{.}\) Let us assume that \(\lambda > 0\text{.}\) 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.

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

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

\begin{equation*}
X(x) = A e^{\omega x} + B e^{-\omega x} + C \sin (\omega x) +
D \cos (\omega x) .
\end{equation*}

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

\begin{equation*}
X(x) = A e^{\omega x} - A e^{-\omega x} + C \sin (\omega x) .
\end{equation*}

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

Now \(T'' + n^4 \pi^4 a^4 T = 0\text{.}\) The general solution is \(T(t) = A \sin (n^2 \pi^2 a^2 t) + B \cos (n^2 \pi^2 a^2 t)\text{.}\) 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 \(T_n(t) = \cos (n^2 \pi^2 a^2 t)\text{.}\)

As 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

\begin{equation*}
f(x) = \sum_{n=1}^\infty b_n \sin (n \pi x) .
\end{equation*}

Then the solution to (5) is

\begin{equation*}
y(x,t) = \sum_{n=1}^\infty b_n
X_n(x) T_n(t)
= \sum_{n=1}^\infty b_n
\sin (n \pi x) \cos ( n^2 \pi^2 a^2 t ) .
\end{equation*}

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 \(T_n(0) = 1\text{,}\) we have

\begin{equation*}
y(x,0) = \sum_{n=1}^\infty b_n X_n(x) T_n(0) =
\sum_{n=1}^\infty b_n X_n(x) =
\sum_{n=1}^\infty b_n
\sin (n \pi x) = f(x) .
\end{equation*}

So \(y(x,t)\) solves (5).

The natural (angular) frequencies of the system are \(n^2 \pi^2 a^2\text{.}\) These frequencies are all integer multiples of the fundamental frequency \(\pi^2 a^2\text{,}\) so we get a nice musical note. The exact frequencies and their amplitude are what musicians call the *timbre* of the note (outside of music it is called the spectrum).

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, \(1,2,3,4,5,\ldots\text{.}\) For a steel beam we get only the square multiples \(1,4,9,16,25,\ldots\text{.}\) 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.

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

\begin{equation*}
f(x) = \sum_{\substack{n=1\\n \text{~odd}}}^\infty \frac{4}{5\pi^3 n^3}
\sin (n \pi x) .
\end{equation*}

Hence, the solution to (5) with the given initial position \(f(x)\) is

\begin{equation*}
y(x,t) = \sum_{\substack{n=1\\n \text{~odd}}}^\infty \frac{4}{5\pi^3 n^3}
\sin (n \pi x) \cos ( n^2 \pi^2 a^2 t ) .
\end{equation*}

Suppose you have a beam of length 5 with free ends. Let \(y\) be the transverse deviation of the beam at position \(x\) on the beam (\(0 < x < 5\)). You know that the constants are such that this satisfies the equation \(y_{tt} + 4 y_{xxxx} = 0\text{.}\) Suppose you know that the initial shape of the beam is the graph of \(x(5-x)\text{,}\) and the initial velocity is uniformly equal to 2 (same for each \(x\)) in the positive \(y\) direction. Set up the equation together with the boundary and initial conditions. Just set up, do not solve.

Suppose you have a beam of length 5 with one end free and one end fixed (the fixed end is at \(x=5\)). Let \(u\) be the longitudinal deviation of the beam at position \(x\) on the beam (\(0 < x < 5\)). You know that the constants are such that this satisfies the equation \(u_{tt} = 4 u_{xx}\text{.}\) Suppose you know that the initial displacement of the beam is \(\frac{x-5}{50}\text{,}\) and the initial velocity is \(\frac{-(x-5)}{100}\) in the positive \(u\) direction. Set up the equation together with the boundary and initial conditions. Just set up, do not solve.

Suppose the beam is \(L\) units long, everything else kept the same as in (5). What is the equation and the series solution?

Suppose you have

\begin{equation*}
\begin{aligned}
& a^4 y_{xxxx} + y_{tt} = 0 \quad (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), \quad y_{t}(x,0) = g(x) .
\end{aligned}
\end{equation*}

That is, you have also an initial velocity. Find a series solution. Hint: Use the same idea as we did for the wave equation.

Suppose you have a beam of length 1 with hinged ends. Let \(y\) be the transverse deviation of the beam at position \(x\) on the beam (\(0 < x < 1\)). You know that the constants are such that this satisfies the equation \(y_{tt} + 4 y_{xxxx} = 0\text{.}\) Suppose you know that the initial shape of the beam is the graph of \(\sin (\pi x)\text{,}\) and the initial velocity is 0. Solve for \(y\text{.}\)

Answer

\(y(x,t) = \sin(\pi x) \cos (4 \pi^2 t)\)

Suppose you have a beam of length 10 with two fixed ends. Let \(y\) be the transverse deviation of the beam at position \(x\) on the beam (\(0 < x < 10\)). You know that the constants are such that this satisfies the equation \(y_{tt} + 9 y_{xxxx} = 0\text{.}\) Suppose you know that the initial shape of the beam is the graph of \(\sin(\pi x)\text{,}\) and the initial velocity is uniformly equal to \(x(10-x)\text{.}\) Set up the equation together with the boundary and initial conditions. Just set up, do not solve.

Answer

\(9 y_{xxxx} + y_{tt} = 0 \quad (0 < x < 10, t > 0)\text{,}\) \(y(0,t) = y_{x}(0,t) = 0\text{,}\) \(y(10,t) = y_{x}(10,t) = 0\text{,}\) \(y(x,0) = \sin(\pi x), \quad y_{t}(x,0) = x(10-x)\text{.}\)