Destabilization paradox due to breaking the Hamiltonian and reversible symmetry
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Citations
Signatures of three coalescing eigenfunctions
Signatures of three coalescing eigenfunctions
Paradoxes of dissipation-induced destabilization or who opened Whitney's umbrella?
Paradoxes of dissipation-induced destabilization or who opened Whitney's umbrella?
Brake comfort – a review
References
Geometrical Methods in the Theory of Ordinary Differential Equations
Treatise on Natural Philosophy
Introduction to the theory of stability
Related Papers (5)
Dissipation-induced instabilities in finite dimensions
Effects of damping on mode-coupling instability in friction induced oscillations
Frequently Asked Questions (17)
Q2. What is the reason for the loss of stability of a gyroscopic system?
Dihedral angle singularity is responsible for the loss of stability by a gyroscopic system, which is statically stable in the absence of gyroscopic forces, due to action of the small damping and circulatory forces.
Q3. How can the asymptotic stability of a circulatory system be reached?
For indefinite matrices D violating inequality (26), the asymptotic stability can be reached only in the presence of gyroscopic, damping, and circulatory forces.
Q4. What is the domain of asymptotic stability?
The domain of asymptotic stability is twisting around the Ω-axis in such a manner that it always remains in the half-space δ > 0, Fig. 10(a).
Q5. What is the apical angle of the asymptotic stability domain of system?
the asymptotic stability domain of system (1) in the space (δ, ν, Ω) near the ν-axis looks like a dihedral angle which becomes more acute while approaching the points ±νf .
Q6. What is the definition of the critical gyroscopic parameter cr?
they describe in an implicit form a limit of the critical gyroscopic parameter Ωcr(δ, γδ) when δ tends to zero, as a function of the ratio γ = ν/δ, Fig. 9(b).
Q7. What is the boundary of the asymptotic stability domain of the unperturbed?
In case when the stability domain of the unperturbed circulatory system has a common boundary with the divergence domain, as shown in Fig. 1(a), the boundary of the asymptotic stability domain of the perturbed system (1) possesses the trihedral angle singularity at ν = ±νd.
Q8. What is the effect of small damping and nonconservative positional forces on the stability?
For a general linear mechanical system with two degrees of freedom the effect of small damping and nonconservative positional forces on the stability of a gyroscopic system as well as the effect of small gyroscopic and damping forces on the stability of a circulatory system has been studied.
Q9. What is the shape of the stability domain at 0?
To study the shape of the stability domain at Ω → 0 the authors note that|2trKD − trKtrD| − |trD| √ (trK)2 − 4 detK ≤ 0, (50)if D satisfies condition (26).
Q10. What is the boundary of the asymptotic stability domain of a multiparameter?
As it has been established by Arnold [3], the boundary of the asymptotic stability domain of a multiparameter family of real matrices is not a smooth surface.
Q11. What is the boundary of the asymptotic stability domain of the perturbed system?
The present paper shows that in both cases the boundary of the asymptotic stability domain of the perturbed system possesses singularities such as “Dihedral angle” and “Whitney umbrella” that govern stabilization and destabilization.
Q12. What is the gyropendulum's domain of asymptotic?
the system stable at Ω = 0 can become unstable at greater Ω, asAc cept edm anus crip tFigure 11: The Crandall gyropendulum and its domain of asymptotic stability.
Q13. What is the gyroscopic stability of a statically unstable conservative system?
the domain of asymptotic stability given by inequalities (15) and (20) consists of two pockets of two Whitney umbrellas, selected by the condition δtrD > 0. Equations (51) are a linear approximation to the stabilityAc cept edm anus crip tFigure 9: Blowing the domain of gyroscopic stabilization of a statically unstable conservative system with K < 0 up to the domain of asymptotic stability with the Whitney umbrella singularities (a).
Q14. What is the difference between the stationary and the rotating damping?
Contrary to the stationary damping, which is a velocity-dependent force, the rotating one is also proportional to the displacements by a non-conservative way and thus contributes not only to the matrix D in equation (1), but to the matrix N as well.
Q15. What is the simplest way to get an impression of the behavior of the functions 0?
To get an impression of the behavior of the functions ν±0 (β), the authors calculate and plot them, normalized by νf , for the following positive-definite matrix K and indefinite matrix D = Di, where i = 1, 2, 3K =⎛ ⎝ 27 33 5⎞ ⎠ , D1 = ⎛ ⎝ 6 33 1⎞ ⎠ , D2 = ⎛ ⎝ 7 43√130 − 114 3√ 130 − 11 1⎞ ⎠ , D3 = ⎛ ⎝ 7 55 1⎞ ⎠ . (30)The graphs of the functions ν±0 (β) bifurcate with a change of detD.
Q16. What is the interesting situation for many applications?
The most interesting for many applications is the situation when system (1) is close either to circulatory system (2) with δ, Ω ν (imperfect reversible system) or to gyroscopic system (3) with δ, ν Ω (imperfect Hamiltonian system).
Q17. What is the coefficient of the equations of motion linearized about the non-inverted state?
The equations of motion linearized about the non-inverted state with the spin rate γ are in the form of themodified Maxwell-Bloch equations (59) with the coefficientsδ = − (1 + e) 2η −1 + e2μ, Ω = σ −1 + e2μ, κ = Fr−1eμ −1 + e2μ, ν = η(1 + e) −1 + e2μ, (71)where the dimensionless inertia ratio σ, Froude number Fr, mass μ, and friction coefficient η areσ =