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Destabilizability of linear Hamiltonian systems. (Russian) Zbl 0609.34055
The authors discuss the stability of the zero solution of the so called linear Hamiltonian system (1) \(\dot x=A(t)x\), \(x\in {\mathbb{R}}^{2n}\), \(t\in {\mathbb{R}}^+\); where A(t) is a piecewise continuous, bounded linear operator. Let \(J: {\mathbb{R}}^ n\to {\mathbb{R}}^ n\) be an operator satisfying \(J^{-1}=J^*=-J\). The system (1) is called linear Hamiltonian if JA(t) is a selfadjoint operator for any \(t\in {\mathbb{R}}^+.\)
Let \({\mathcal H}\) denote the set of such systems with the sup norm \(\| A\| =\sup_{t>0}\| A(t)\| =\sup_{t>0}\{\sup_{| x| =1}(| A(t)\cdot x|)\}\) and \(\dot {\mathcal H}^ a \)periodic subset of \({\mathcal H}\). The authors recall the following result of V. I. Arnol’d. A system \(A\in \dot {\mathcal H}\) is linear Hamiltonian if and only if its Cauchy operator X(t,s), \(t,s\in {\mathbb{R}}^+\) is symplectic, that is \(J^*X^*JX'=I\) (the identity operator). The system A is said to be symplecticly transformable into system B if there exists a symplectic Lyapunov transformation \(y=L(t)x\) transforming \(\dot x=A(t)x\) into \(\dot y=B(t)y.\)
The authors prove a number of interesting results, such as: A system \(A\in {\mathcal H}\) is stable if and only if it can be symplectically transformed into the system \(\dot y\equiv 0\). In the space \({\mathcal H}\) the stable systems form a nowhere dense set. If \({\mathcal H}_ 0(A)\) denotes the subset of systems \(B\in {\mathcal H}\) such that \(\lim_{t\to \infty}\| B(t)-A(t)\| =0,\) then the following theorem is true: For any stable system \(A\in {\mathcal H}\) there exists an unstable system \(B\in {\mathcal H}_ 0(A)\).
Reviewer: V.Komkov

MSC:
34D20 Stability of solutions to ordinary differential equations
34A30 Linear ordinary differential equations and systems
70H05 Hamilton’s equations
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