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Peter Lax showed that the differential operators $$L=-6\partial_x^2-u,\quad B=-4\partial_x^3-u\partial_x-(1/2)u_x$$ fulfilling the Lax equation $$\dot{L}+[L,B]=0$$ is equivalent to the KdV equation $$u_t+uu_x+u_{xxx}=0.$$

For the harmonic oscillator one knows a Lax representation given by two two-by-two matrices.

But: Are there differential operators $L$ and $B$ which are equivalent to the harmonic oscillator equation $$\ddot{q}+\omega q=0$$ and fulfilling the Lax equation? If yes: How can one derive the form of these operators?

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  • $\begingroup$ Do you mean that Lax proved $(\dot{L}+[L,B])u=0 \Leftrightarrow u_t+u u_x + u_{xxx}=0$? (in particular, the $x$ is taken to be independent from time? i.e. $\partial_t (\partial^2_x)=\partial^2_x\partial_t$ or not?) However this is a PDE while the harmonic oscillator equation is an ODE. $\endgroup$
    – yuggib
    May 7, 2015 at 7:46
  • $\begingroup$ Yes that is what I meant. Besides the fact that the harmonic oscillator is an ODE, is there no way to write down such matrices? Yes x is independent of time, so everything commutes. $\endgroup$
    – Richard
    May 7, 2015 at 7:52
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    $\begingroup$ Well, I am not sure if I understand completely, because $L=\partial_t$ and $B=\omega t$ (where $\omega\in\mathbb{R}$ does not depend on time) seem to do what you are asking...I will use the symbol $\partial$ here even if there is only $t$ as a variable, so it is not so meaningful (but shorter to write): $\partial_t L q=\partial_t^2 q$; $LBq=\omega q+\omega t\partial_t q$, $BLq=\omega t\partial_t q$, i.e. $[L,B]q=\omega q$. $\endgroup$
    – yuggib
    May 7, 2015 at 9:21

1 Answer 1

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Well, consider a Lax pair for the Harmonic Oscillator; \begin{equation} L = \begin{pmatrix} p & \omega q \\ \omega q & -p \\ \end{pmatrix}, \quad M=\frac{\omega}{2} \begin{pmatrix} 0 & -1 \\ 1& 0 \\ \end{pmatrix} \end{equation} Since the Hamiltonian is $$H(q, p)=\frac{1}{2}(p^{2}+\omega^{2}q^{2})$$ It is eay to check that the Lax equation $\dot{L}=[M, L ]$ is equivalent to the Hamiltonian system \begin{equation} \dot{q}=\frac{\partial H}{\partial p}=p, \quad \dot{p}=-\frac{\partial H}{\partial q}=-\omega^{2}q \end{equation} It is a trivial exercise to show that \begin{equation} \ddot{q}+\omega^{2}q=0 \end{equation}

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