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I have doubts about how to build the plot of dispersion relation for a general finite phononic system, in particular spring-mass systems. Eqs of any of these systems can be written in matrix form as $$ D\vec{u}=\omega^2 \vec{u} \tag{1} $$ For the case of "infinite" system, usually one set the periodic boundary condition (PBC) and under the Bloch expansion $u_i=u_0 \mathrm{e}^{ikn}$ in (1), one can solve analytically $\omega(k)$ for a single primitive cell. As example, in the 1d diatomic chain one can obtain $\omega(k)$ by this way

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In this case, considering $N$ masses and the PBC, this quantizes the values of $k$, yielding $N$ points in $\omega(k)$, but taking $N\to\infty$, one can plot a continuous line in $\omega(k)$ (similar for monoatomic chain).

However in a finite system, under specific boundary conditions in $D$, if I impose the Bloch expansion in (1), the lack of translational invariance of $D$ doesn't allow us to reduce (1) to the dynamics of a single cell. I'm not interested in analytically solving $\omega(k)$ in these system, but I would like to be able to plot the N points of $\omega(k)$. I think that I'm wrong imposing $u_0 \mathrm{e}^{ikn}$ here, but then I don't know how to make appear the wave vector $k$ in (1) to obtain the $\omega(k)$ relations. Is there a generic numerical way to obtain the points of $\omega(k)$ in these systems?

(e.g. the diatomic chain with one end free and the other fixed)

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You can still use periodic boundary conditions for a finite chain if the ends are fixed (as I will demonstrate below), and so the dispersion relation is unchanged in the sense that you just replace the continuous variable $k$ with its discrete version.

To treat a finite chain first impose periodicity as you did before, this means that the two ends of the chain $x_0$ and $x_N$ will be regarded as the same point but this is okay since we fix both ends by demanding that $$x_0(t) = x_N(t) = 0$$ If we look at the general solution $$x_n(t) = \sum_{k=-\pi/a}^{\pi/a}e^{ikna}(A_ke^{i\Omega_k t}+B_ke^{-i\Omega_k t})$$where $\Omega_k$ is the mode frequency, the ends are fixed if we set $A_k=-A_{-k}$, $~B_k=-B_{-k}$, and $A_0=B_0=0$.

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  • $\begingroup$ Thanks for your answer, but what would happen if I wanted to leave one end fixed and the other free? $\endgroup$
    – FVergara
    Commented Feb 17, 2022 at 18:13

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