Old stuff, to be rewritten. See the source in anotacioneslatex.tex.
They are important physical system what can be modeled, by approximation in their lagrangian to second order Taylor expansion, by
$$ M\ddot{X}=-K X $$where $M$ and $K$ are symmetric matrices ($M$ comes from kinetic energy in the lagrangian and $K$ is the Hessian matrix of a potential function $V$ also in the exact lagrangian).
To solve it, we construct the associated first order system:
$$ Y=\left(\begin{array}{c} X \\ \dot{X} \end{array}\right) $$and then
$$ \dot{Y}=\left(\begin{array}{cc} 0 & I \\ -M^{-1} K &0 \end{array}\right) \cdot Y $$ $$ \dot{Y}=A\cdot Y $$To find eigenvalues and eigenvectors in order to use equation \ref{soluciondefinitiva} we observe that
\begin{itemize}
\item $\lambda$ is eigenvalue of $A$ iff $\lambda^2$ is eigenvalue of $-M^{-1}K$. This is so because the block matrix determinant formula
$$ det\left(\begin{array}{cc} H & B \\ C & D \end{array}\right)=det(H-BD^{-1}C)\cdot det(D)$$applied to $\lambda I-A$.
is an eigenvector of $A$ associated to $\lambda$ then $\psi$ is an eigenvector of $-M^{-1}K$ associated to $\lambda^2$. This is so because
$$ A\xi=\left(\begin{array}{cc} 0 & I \\ -M^{-1} K &0 \end{array}\right) \cdot \left( \begin{array}{c} \psi \\ \eta \end{array}\right)=\left( \begin{array}{c} \eta \\ -M^{-1}K\psi \end{array}\right)=\left( \begin{array}{c} \lambda \psi \\ \lambda \eta \end{array}\right) $$So,
$$ A^2\xi=\left(\begin{array}{cc} -M^{-1} K & 0 \\ 0 &-M^{-1} K \end{array}\right) \cdot \left( \begin{array}{c} \psi \\ \eta \end{array}\right)=\left( \begin{array}{c} -M^{-1} K \psi \\ -M^{-1} K \eta \end{array}\right)= \left( \begin{array}{c} \lambda^2\psi \\ \lambda^2 \eta \end{array}\right) $$And therefore
$$ -M^{-1} K \psi =\lambda^2\psi $$Reciprocally, if $(\delta, \psi)$ is a pair such that $- M ^ { - 1 } K \psi = \delta \psi$ (of course $\delta$ is real and negative) then is easy to check that we have two eigenvalues with two eigenvectors for
$$ \left(\begin{array}{cc} 0 & I \\ -M^{-1} K &0 \end{array}\right) $: the pair $(\sqrt{\delta},\left( \begin{array}{c} \psi\\ \sqrt{\delta} \psi \end{array}\right))$ and the pair $(-\sqrt{\delta},\left( \begin{array}{c} \psi\\ -\sqrt{\delta} \psi \end{array}\right)$$\end{itemize}
All the eigenvalues of $-M^{-1} K$ are reals, since is a symmetric matrix and they always can be diagonalized. Moreover, they are all negative numbers, because this matrix reflect a stable equilibrium point, i.e., from it potential energy increases in whatever direction we take. So we conclude:
\begin{itemize}
\item All the eigenvalues are pure imaginary ones. There is no $e^{\alpha t}$ terms.
\item For every eigenvalue of $A$ it can be chosen an eigenvector
$$ \psi=\left( \begin{array}{c} \xi \\ \eta \end{array}\right) $$ such that $\xi$ is real (because is an eigenvector of $-M^{-1} K$ and their eigenvalues are all real). So $$ Im(\psi)=\left( \begin{array}{c} 0 \\ Im(\eta) \end{array}\right). $$\end{itemize}
So for every eigenvalue-eigenvector of $-M^{-1}K$, $(\lambda_n,\xi)$, we have two solutions:
$$ \Psi^{n}=e^{i\omega_n t} \xi^n $$ $$ \tilde{\Psi}^{n}=e^{-i\omega_n t} \xi^n $$where $\omega_n=\frac{\sqrt{\lambda_n}}{i} \in \RR$ (remember $\lambda_n$ is negative). It may seem that this both solutions have the same initial value, which is a contradiction, but they have different initial velocities!
Moreover, observe that$
(\tilde{\Psi}^{n})^*=\Psi^n
$, because we can choose $\xi^n$ to be real.
So if we look for \textbf{real solutions} we have that
$$ \Psi(t)=\sum_n C_n \Psi^n+\tilde{C}_n\tilde{\Psi}^n $$ $$ (\Psi(t))^*=\sum_n C_n^* (\Psi^n)^*+(\tilde{C}_n)^*(\tilde{\Psi}^n)^* $$must be equal, and since the solutions form a vector space we conclude that $C_n^*=\tilde{C}_n$
and so
where $A_n$ and $\phi_n$, determined by initial conditions, are reals and have absorbed the constant to leave $\xi^n$ as a real vector.
\bigbreak
These fundamental solutions are called \textbf{normal modes}. The constants $\omega_k$ are called \textbf{fundamental frecuencies.}
\begin{center}
\includegraphics[width=15cm]{imagenes/noscillators.png}
\end{center}
We are going to study $N$ masses connected to a string with tension. We leave open ends, for the moment.
Let $\Psi_j (t)$ be the displacement from the equilibrium of every mass $j$. Let the mass be $m$ and the distance between the mass be $d$. Sometimes we will take $x=jd$ and write
$$ \Psi(x,t)=\Psi_{x/d} (t) $$Analyzing every mass individually we obtain
$$ \ddot{\Psi}_1=-\frac{2T}{md}\Psi_1+\frac{T}{md}\Psi_2 $$ $$ \ddot{\Psi}_2=\frac{T}{md}\Psi_1-\frac{2T}{md}\Psi_2+\frac{T}{md}\Psi_3 $$In general:
$$ \ddot{\Psi}=\frac{1}{md}\left( \begin{array}{cccc} {- T} & {T} & {0} & {0} \\ {T} & {-2 T} & {T} & {0} \\ {...} & {...} & {...} & ...\\ {0} & {0} & {T} & {-T} \end{array}\right) \cdot \Psi $$or in short
$$ \ddot{\Psi}=-M^{-1}K\cdot \Psi $$Observe that the matrix $-M^{-1}K$ is symmetric and all their eigenvalues are negatives, since it comes from a stable equilibrium point (see section \ref{coupledoscillators}). Moreover, from this section and the previous ones we know that a basis for solutions of this differential equation is
$$ A cos(\omega t+ \phi) \psi $$where $-\omega^2$ is an eigenvalue of $-M^{-1}K$,
$\psi$ is an eigenvector for this eigenvalue, and $A$ and $\phi$ are real constants determined by the initial conditions: the $A$'s by the initial positions and the $\phi$'s by the initial velocities.
\bigbreak
\textbf{When $N$ is really big} is very difficult to find the eigenvalues by linear algebra, so we can proceed in a different way.
Also, the vector $\psi$ is a very large vector, since the number of masses $N$ will tend to infinity, and we want to study the shape'' of their components. In fact, when $N$ goes to infinity $\psi$ will be a function of position, giving the displacement of the masses in the initial time.
As usual, we will follow with the complex solution for $\Psi$ and take the real part in the final step. We think in solutions like
where $\xi$ is a complex eigenvector of infinite length (keep an eye: in section \ref{coupledoscillators} we saw that with finite masses we can force $\xi$ to be real, but with infinite masses, a priori, we cannot).
To find infinite length eigenvectors we use the trick of \ref{symmetrytrick}. Since this infinite chain of oscillating masses has translation symmetry, we check that
$$ -M^{-1}K S=S (-M^{-1}K) $$where
$$ -M^{-1}K= \left(\begin{array}{cccccc}{\ddots} & {\vdots} & {\vdots} & {\vdots} & {\vdots} & {\ddots} \\ {\cdots} & {-2 B} & {C} & {0} & {0} & {\cdots} \\ {\cdots} & {C} & {-2 B} & {C} & {0} & {\cdots} \\ {\cdots} & {0} & {C} & {-2 B} & {C} & {\cdots} \\ {\cdots} & {0} & {0} & {C} & {-2 B} & {\cdots} \\ {\ddots} & {\vdots} & {\vdots} & {\vdots} & {\vdots} & {\ddots}\end{array}\right) $$and
$$ S=\left(\begin{array}{cccccc}{\ddots} & {\vdots} & {\vdots} & {\vdots} & {\vdots} & {\ddots} \\ {\cdots} & {0} & {1} & {0} & {0} & {\cdots} \\ {\cdots} & {0} &0& {1} & {0} & {\cdots} \\ {\cdots} & {0} & {0} & 0 &{1} & {\cdots} \\ {\cdots} & {0} & {0} & {0} & 0 & {\cdots} \\ {\ddots} & {\vdots} & {\vdots} & {\vdots} & {\vdots} & {\ddots}\end{array}\right) $$What are the eigenvectors for $S$? We can take any $\beta \in \RR$, and then produce an eigenvector $\xi$. Observe that
$$ S\xi (j)=\xi(j+1)=\beta \xi(j) $$so if we take $\xi(0)=1$ we conclude $\xi(j)=\beta^j$.
Moreover, since $j\in \ZZ$ we conclude that $|\beta|=1$ and therefore must be $\alpha \in [-\pi, \pi]$ such that $\beta=e^{i\alpha}$. In conclussion, for every $\alpha \in [-\pi, \pi]$ we have the eigenpair $(e^{i\alpha},\xi^{\alpha})$ where $\xi^{\alpha}(j)=e^{i\alpha j}$, respect to $S$.
But, what is the eigenvalue respect to $-M^{-1}K$?
So $\xi^{\alpha}$ is an eigenvector of $-M^{-1}K$ associated to the eigenvalue $2C\cos(\alpha)-2B$, which must be negative (probably $C=B$, and we would get that).
Now, think that the pair $(2C\cos(\alpha)-2B,\xi^{\alpha})$ produces two solutions for the associated first order system (and then truncated''):
where $\omega_{\alpha}=\frac{\sqrt{2C\cos(\alpha)-2B}}{i} \in \RR$. In the particular case where $B=C=\frac{T}{md}$,
$$ \omega_{\alpha}=2\sqrt{\frac{T}{md}}|\sin{\frac{\alpha}{2}}| $$which is known as \textit{dispersion relation}
\bigbreak
A few remarks:
\begin{itemize}
\item The general solution is a linear combination of $\Psi^{\alpha}$'s and $\tilde{\Psi}^{\alpha}$'s. But since $\alpha$ has no restriction, beyond $\alpha \in [-\pi, \pi]$, because we don't have boundary conditions, $\alpha$ varies continuously. So, instead of a linear combination we get
$$ \Psi=\int_{-\pi}^{\pi} C(\alpha) \Psi^{\alpha} +\tilde{C}(\alpha) \tilde{\Psi}^{\alpha}d \alpha $$Observe that this has been developed with a trick: we forgot the initial and final rows of the matrix $-M^{-1}K$ since we are dealing with a big $N$. This is the reason why we are finding more than $N$ eigenvalues (infinite, actually): since we are no taking into account the initial and final restriction, it is totally as if we had infinite masses.
\item Let's look for the general \textbf{real} solution. First, observe that:
Then
$$ \Psi^*=\int_{-\pi}^{\pi} C^*(\alpha) (\Psi^{\alpha})^* +\tilde{C}^*(\alpha) (\tilde{\Psi}^{\alpha})^*d \alpha= \int_{-\pi}^{\pi} C^*(\alpha) \tilde{\Psi}^{-\alpha} +\tilde{C}^*(\alpha) {\Psi}^{-\alpha}d \alpha= $$ $$ =\int_{-\pi}^{\pi} C^*(-\alpha) \tilde{\Psi}^{\alpha} +\tilde{C}^*(-\alpha) {\Psi}^{\alpha}d \alpha $$Since it must be $\Psi=\Psi^*$, and so :
$$ C^*(-\alpha)=\tilde{C}(\alpha) $$(I don't know how to prove this!!, but the idea is that $\Psi^{\alpha}$ and $\tilde{\Psi}^{\alpha}$ constitute a basis for the vector space of solutions) and therefore:
$$ \Psi=\int_{-\pi}^{\pi} C(\alpha) \Psi^{\alpha} +\tilde{C}(\alpha) \tilde{\Psi}^{\alpha}d \alpha=\int_0^{\pi} 2Re[C(\alpha) \Psi^{\alpha} +\tilde{C}(\alpha) \tilde{\Psi}^{\alpha}]d\alpha $$\item The parameter $\alpha$ is the seed for the wave number. Let $d$ be the distance between the masses, we can write $x=d\cdot j$ to be the position in the horizontal direction. If we try to write everything int terms of $x$ instead $j$ is useful to choose $\kappa=\frac{\alpha}{d}$ for the trial solution for the eigenvector. The normal modes would be:
$$ \Psi^{\kappa}(x, t)=a \cdot cos(\omega_{\kappa} t+\kappa x+\phi) $$\item Dispersion relation is a name for the functional relation between wave number $\kappa$ and frecuency $\omega_{\kappa}$. In this case is
$$ \omega_{\kappa}=2\sqrt{\frac{T}{md}}sin(\frac{\kappa d}{2 }) $$\item The idea for choosing $\xi(j)=A e^{i \alpha j}$ can be seen from other point of view.
The eigenvectors satisfy the relation
\begin{equation}\label{eigenvectorrelation}
\end{equation}
where we forget the first and the last relation because we assume $N$ very very large
One can observe that if $d$ is very small and taking into account the name change $x=jd$, equation \ref{eigenvectorrelation} can be interpreted (approximately) as:
$$ - m\omega^2 \xi(x)=T\left[ \frac{\xi(x+d)-\xi(x)}{d}-\frac{\xi(x)-\xi(x-d)}{d}\right] \approx T [\xi'(x+d/2)]-\xi'(x-d/2)] $$and so
$$ - m\omega^2 \xi(x)\approx T d \xi''(x) $$But since $m=d\mu$, where $\mu$ is the density and is supposed to be constant
$$ - \frac{\mu \omega^2}{T} \xi(x)\approx \xi''(x) $$We have got an equation quite similar but in the variable $x$ or $j$ if you prefer, so is natural to choice $\xi(j)=A e^{i \alpha j}$.
\end{itemize}
\section{Boundary conditions}
We have studied an infinite system. If we want to come back to a finite system we impose boundary conditions, so we reduce the number of fundamental solutions that are allowed.
Particular case: $N=4$. We have:
$$ \ddot{\Psi}=\frac{1}{md}\left( \begin{array}{cccc}{- T} & {T} & {0} & {0} \\ {T} & {-2 T} & {T} & {0} \\ {0} & {T} & {-2T} & T\\ {0} & {0} & {T} & {-T}\end{array}\right) \cdot \Psi $$We solve using Mathematica and find, evidently, four eigenvalues:
and four eigenvectors
$$ \xi_1=(-1,1+\sqrt{2},-1-\sqrt{2}, 1), \xi_2=(1,-1,-1,1), $$With pictures ($T=2 ; m=1 ; d=0.5$):
\begin{tabular}{cc}
\includegraphics[width=60mm]{imagenes/sol1.png} &
\includegraphics[width=60mm]{imagenes/sol2.png} \\
\includegraphics[width=60mm]{imagenes/sol3.png} &
\includegraphics[width=60mm]{imagenes/sol4.png} \\
\end{tabular}
Could we obtain the same with the big-N-technique? Let's see. We start with infinite vibrating masses with displacement $\Psi_j(t)$, and so infinite fundamental frequencies
$$ \omega_{\alpha}=2 \sqrt{\frac{T}{m d}} \sin (\alpha / 2) $$one for every $\alpha \in [0,\pi]$.
If we want to restrict to this particular case, we have to impose some conditions: \textbf{boundary conditions}. For example, if we force $\Psi_0=\Psi_1$ and $\Psi_4=\Psi_5$ we are removing the effect of masses 0 and 5 over the masses ranging from 1 to 4 (as desired). Let's study this two conditions: we are going to impose $\Psi_0=\Psi_1$ and $\Psi_4=\Psi_5$ and watch what $\alpha$'s survive.
$$ e^{i\omega_{\alpha}t} [ C(\alpha)+C(-\alpha)]=e^{i\omega_{\alpha}t} [ C(\alpha) e^{i\alpha}+C(-\alpha) e^{-i\alpha}] $$ $$ C(\alpha)+C(-\alpha)= C(\alpha) e^{i\alpha}+C(-\alpha) e^{-i\alpha} $$and together with
$$ e^{i\omega_{\alpha}t} [ C(\alpha) e^{i\alpha 4}+C(-\alpha) e^{-i\alpha 4}]=e^{i\omega_{\alpha}t} [ C(\alpha) e^{i\alpha 5}+C(-\alpha) e^{-i\alpha 5}] $$we arrive to
$$ \alpha=\frac{\pi n}{4} $$Let's study this values one by one:
\begin{enumerate}
\item n=0. Then $\alpha=0$, $\omega_{\alpha}^2=0$ and $\xi=(1,1,1,1)$.
Correspond to the case $\omega_{4}$ following Mathematica.
\item $n=1$. Then $\alpha=\frac{\pi}{4}$, and $\omega_{\alpha}^2=\frac{T}{md}(2-\sqrt{2})$. We are in case 3 from Mathematica. The eigenvector obtained is $\xi=(e^{i\frac{\pi}{4}},e^{i\frac{\pi}{2}},e^{i\frac{3\pi}{4}},e^{i\pi})$, which is complex and do not coincide with the one obtained by Mathematica. But since $-\alpha$ produces the same eigenvalue, we can mix the previous eigenvector with $(e^{-i\frac{\pi}{4}},e^{-i\frac{\pi}{2}},e^{-i\frac{3\pi}{4}},e^{-i\pi})$ to make new eigenvectors for this eigenvalue. In fact is a (long) computation to check that the linear combination of this complex vectors that produces the same that the real above eigenvector is
$$ \left\{\left(x \rightarrow-\frac{1}{2}+i\left(-\frac{1}{2}+\frac{1}{\sqrt{2}}\right), y \rightarrow-\frac{1}{2}+i\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)\right\}\right. $$I found it by using Mathematica.
\item $n=2$. Then $\alpha=\frac{\pi}{2}$, and $\omega_{\alpha}^2=2\frac{T}{md}$. This correspond with case $\omega_2$ above. The eigenvector is not the same, but the phenomenom is the same that for $n=1$: we can recover by mixing both eigenvectors corresponding to $\alpha$ and $-\alpha$.
\item $n=3$. Then $\alpha=\frac{3\pi}{4}$, and $\omega_{\alpha}^2=\frac{T}{md}(2+\sqrt{2})$. Idem.
\item $n=4$. Then $\alpha=\pi$. Condition $\Psi_0^{\pi}=\Psi_1^{\pi}$ implies $C(\pi)+C(-\pi)=0$. Moreover, since $e^{i\pi j}=e^{-i\pi j}$, we deduce that $C(\pi)\Psi^{\pi}+C(-\pi)\Psi^{-\pi}=0$, so we can ignore this normal mode.
\item $n=5$. Then $\alpha=\pi+\frac{\pi}{4}$. As we reasoned above, this would be the same case as $\pi+\frac{\pi}{4}-2\pi=-\frac{3 \pi}{4}$ which, in fact, is paired with $\alpha=\frac{3\pi}{4}$ (case $n=3$)
\item All the remaining cases are repeated
\end{enumerate}
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Author of the notes: Antonio J. Pan-Collantes
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