Exact invariant covariance and stability threshold for scalar harmonic BAOAB

The canonical density factorizes over $q$ and $p$. By [F4] its normalizer is the product of the two one-dimensional Gaussian integrals, and [F5] gives $Z_q=\sqrt{2\pi \mathrm{/}(\beta k)}$ and $Z_p=\sqrt{2\pi m\mathrm{/}\beta }$. The normalized factors are centered. For $c>0$, [F6] on $[-R,R]$ gives ${\int }_{-R}^R{x}^2{e}^{-c{x}^2\mathrm{/}2}dx=\frac{1}{c}({\int }_{-R}^R{e}^{-c{x}^2\mathrm{/}2}dx-2R{e}^{-c{R}^2\mathrm{/}2})\mathrm{.}$ Letting $R\to \mathrm{\infty }$ and using $R{e}^{-c{R}^2\mathrm{/}2}\to 0$ gives ${\int }_{\mathbb{R}}{x}^2{e}^{-c{x}^2\mathrm{/}2}dx=c^{-1}{\int }_{\mathbb{R}}{e}^{-c{x}^2\mathrm{/}2}dx$; oddness gives zero mean. Therefore the canonical variances are $1\mathrm{/}(\beta k)$ and $m\mathrm{/}\beta$, with zero cross-covariance. These are the canonical values used below.

Set $v=p\mathrm{/}m$, $s=h\mathrm{/}2$, and ${\sigma }^2={\beta }^{-1}(1-a^2)\mathrm{/}m$. Applying [F2] in sequence and using $F(q)=-kq$ gives [F1, F2, F3, given, algebra] $q^{\mathrm{\prime }}=(1-r(1+a))q+s(1+a)v+s\sigma \xi ,$ $v^{\mathrm{\prime }}=-s{\omega }^2(1+a)(1-r)q+(a-r(1+a))v+(1-r)\sigma \xi ,$ where $\xi \sim N(0,1)$. Thus $z^{\mathrm{\prime }}=(q^{\mathrm{\prime }},v^{\mathrm{\prime }}{)}^T=Az+g\xi$, with $A=\left(\begin{array}{cc}1-r(1+a)& s(1+a)\\ -s{\omega }^2(1+a)(1-r)& a-r(1+a)\end{array}\right),g=\sigma \left(\begin{array}{c}s\\ 1-r\end{array}\right)\mathrm{.}$

Direct expansion gives $\det A=a$ and $\mathrm{tr}A=(1+a)(1-2r)$. The characteristic polynomial is ${\lambda }^2-(1+a)(1-2r)\lambda +a$. Since $0<a<1$, for $0<r<1$ its values at $1$ and $-1$ satisfy [F1, step 1.2, algebra] $1-\mathrm{tr}A+\det A=2r(1+a)>0,1+\mathrm{tr}A+\det A=2(1+a)(1-r)>0.$ If its roots are nonreal, they are conjugate with modulus $\sqrt{a}<1$. If they are real, their product $a>0$ makes them have the same sign. For two positive roots, the first displayed product is positive and their product is below one, so neither can be at least one. For two negative roots, the second displayed product is positive and their product is below one, so neither can be at most $-1$. Hence both roots have modulus less than one. At $r=1$, the polynomial has root $-1$. At $r>1$, its value at $-1$ is negative while it tends to positive infinity as $\lambda \to -\mathrm{\infty }$, so it has a root ${\lambda }_u<-1$; its other root is $a\mathrm{/}{\lambda }_u$ and has modulus less than one. Therefore $\rho (A)<1$ exactly when $0<r<1$.

For $0<r<1$, the characteristic roots have modulus less than one by step 2.1. If they differ, Cayley–Hamilton writes $A^n$ as a linear combination of their $n$th powers. If they coincide at $\lambda$, Cayley–Hamilton gives $(A-\lambda I{)}^2=0$, so $A^n={\lambda }^{n}I+n{\lambda }^{n-1}(A-\lambda I)$; for $\mathrm{\mid }\lambda \mathrm{\mid }<1$ this is bounded by $C{\theta }^n$ for any fixed $\mathrm{\mid }\lambda \mathrm{\mid }<\theta <1$. Thus the power-decay hypothesis of [L1] holds. Its discrete Lyapunov covariance in $(q,v)$ coordinates is [F1, F3, L1, step 1.2, step 2.1, algebra] $P=\frac{1}{\beta m}\left(\begin{array}{cc}{\omega }^{-2}& 0\\ 0& 1-r\end{array}\right)\mathrm{.}$ Indeed, substituting $A,g$ from step 1.2 gives $AP{A}^T+g{g}^T=P$: the off-diagonal identity reduces to $s(1-r)[(1+a)(a-1)+(1-a^2)]=0$; the first diagonal identity reduces, after multiplication by $\beta m{\omega }^2$, to $[1-r(1+a){]}^2+r(1+a)[2-r(1+a)]=1;$ and the second diagonal identity, after division by $1-r$ and multiplication by $\beta m$, reduces to $[a-r(1+a){]}^2+r(1+a{)}^2(1-r)+(1-a^2)(1-r)=1.$ Both displayed scalar identities follow by expansion and $r=s^2{\omega }^2$. Thus [L1] gives the unique invariant Gaussian with this covariance. Transforming back by $p=mv$ gives $\mathrm{Var}(Q)=1\mathrm{/}(\beta k)$, $\mathrm{Var}(P)=m(1-r)\mathrm{/}\beta$, and zero covariance.

For $r>1$, step 2.1 gives a real left eigenvector $\mathrm{\ell }$ for ${\lambda }_u<-1$. The innovation vector excites that mode: direct multiplication gives $\det [g,Ag]=-2{\sigma }^{2}s(1-r)\mathrm{\ne }0$, so ${\mathrm{\ell }}^{T}g\mathrm{\ne }0$ (otherwise $g$ would lie in the other right eigenspace and $g,Ag$ would be dependent). If an invariant probability existed, its projection $X={\mathrm{\ell }}^{T}z$ would satisfy $X^{\mathrm{\prime }}={\lambda }_{u}X+\zeta$ with independent nondegenerate centered normal $\zeta ={\mathrm{\ell }}^{T}g\xi$. Its characteristic function would obey $\phi (t)=\phi ({\lambda }_{u}t)e^{-\mathrm{Var}(\zeta )t^2\mathrm{/}2}$. Iterating and using $\mathrm{\mid }\phi \mathrm{\mid }\le 1$ forces $\phi (t)=0$ for every $t\mathrm{\ne }0$. For any probability law of a finite real random variable $X$, [F7] and dominated convergence applied to $e^{itX}\to 1$ with majorant $1$ give $\phi (t)\to 1$ as $t\to 0$, contradicting this forced vanishing. Hence no invariant probability exists for $r>1$. Since $r=h^{2}k\mathrm{/}(4m)$, the threshold is $h=2\sqrt{m\mathrm{/}k}$. The boundary $r=1$ has an eigenvalue $-1$, so [L1] does not give a unique invariant law there. This proves the stated stable, boundary, and unstable cases without making any nonharmonic invariant-law claim.