Control of a Wave Equation with a Dynamic Boundary Condition
Summary (1 min read)
Introduction
- The aim of this paper is to study a wave equation in a bounded one-dimensional medium supplied with a dynamic (or kinetic) boundary condition at one end.
- In that case, the control is the torque applied to one extremity, and the kinetic boundary condition is given by the behavior of the rock-tip contact, which is subject to nonlinear friction.
- The authors establish the framework in which they analyze system (1), and state their well-posedness result.
- The authors also introduce the state space X , V ×H (5) endowed with the product Hilbertian structure.
III. STABILITY ANALYSIS OF THE CLOSED-LOOP SYSTEM
- Next, the authors analyze the stability of the closed-loop system when the feedback is linear.
- Now, the authors derive some sufficient conditions for the energy to decay exponentially.
- By a density-continuity argument, the uniform estimate (24) holds for weak solutions as well – see Proposition 1.
- Using a similar argument, one verifies that the limit does not depend on the sequence {tn}.
IV. NUMERICAL SIMULATIONS
- The authors provide some numerical computations for illustrative purposes.
- The authors discretize (1) using finite elements over space and finite differences over time, on the basis of the functional formulation of the problem.
- Computations also suggest that the proportional feedback is not robust to errors: taking values of q slightly below 0.5 induces unclear situations where an exponential decay is not easily identifiable.
VI. CONCLUSION AND PERSPECTIVES
- The well-posedness of a wave equation with a nonlinear dynamic boundary condition and nonlinear Neuman-type feedback is proved, in a variational framework.
- Using a proportional feedback law, exponential stabilization is achieved under suitable assumptions.
- In the linear case, an infinite-dimensional frequencydomain approach would be interesting to tackle the stabilization problem and obtain sharp conditions.
- Besides, adding an integral action to the feedback law should be considered.
- Also, the experimental device of [8] simulating drilling dynamics may be used to test the model in real experiments.
Did you find this useful? Give us your feedback
Citations
1 citations
References
40 citations
31 citations
22 citations
20 citations
19 citations
Related Papers (5)
Frequently Asked Questions (14)
Q2. What is the way to solve the stabilization problem?
In the linear case, an infinite-dimensional frequencydomain approach would be interesting to tackle the stabilization problem and obtain sharp conditions.
Q3. What is the condition for the converse inequality to hold?
It follows from the boundedness of the weights ρ and µ that there exists a constant Mρ,µ > 0 such that∀[u,v] ∈ X , Γρ,µ(u,v) ≤Mρ,µE(u,v). (15)A sufficient condition for the converse inequality to hold is the existence of C1, C2 > 0 such thatµ(x) ≥ C1 and µ(x)2 − ρ(x)2 ≥ C2 (16)holds for all x in Ω.
Q4. What is the simplest way to identify a dense subset of H?
Because V is a dense subset of H , the latter can be identified as a dense subset of V ′, leading to the classical injection chain V ↪→ H ' H ′ ↪→ V ′, where each space is dense and continuously embedded into the following one.
Q5. What is the condition on q derived in Theorem 2?
Further computations suggest that the condition on q derived in Theorem 2 is nearly sharp as taking q = 0.5 leads to an exponentially unstable system.
Q6. What is the simplest way to prove that u H2?
For all w ∈ V , the authors havea(u,w) + g(v(L))w(L) + 2(v,w)H = (f2,w)H . (36)Since f2 belongs to L2(Ω), taking φ = (φ, 0) in (36), where φ is an arbitrary element of C∞c (Ω), allows us to prove that u ∈ H2(Ω).
Q7. What is the condition F (0) = 0?
Assumption 3. The function F is q-Lipschitz for some q < 1/2. Also, F (0) = 0.The condition F (0) = 0 is quite natural when dealing with friction.
Q8. What is the simplest way to stabilize a closed loop system?
From now on, the authors are interested in stabilizing system (1) using a proportional feedback, i.e.U(t) = −α∂tu(L, t), (19)where α is a positive gain to be tuned.
Q9. What is the proof of the Hadamard theorem?
Evaluating (37) with ρ(x) and then ρ(L−x) yields −∂xu(0)+2v(0) = f2(0) and ∂xu(L) = −g(v(L)), hence X = (u,v) ∈ D(Ag) and Ag(X) + 2X = Y, which concludes the proof.
Q10. What is the simplest way to prove that a space is a reflexive space?
reformulating (33), the authors seek v ∈ V such thatΘ(v) = L in V ′. (35)The authors wish to apply Lemma 3 given in Appendix to Θ. As a Hilbert space, V is reflexive; it is also separable – see Lemma 1.
Q11. What is the simplest way to denote the 'h' subset of H?
define an unbounded (nonlinear) operator Ag on X by D(Ag) , {[u,v] ∈W × V : ∂xu(L) = −g(v(L))}∀X = [u,v] ∈ D(Ag), Ag(X) , − [v (∂xxu, ∂xu(0))] .(26) Note that the domain D(Ag) need not be a subspace.
Q12. What is the definition of a distributional identity?
With an integration by parts, and using the definition of D(Ag), the authors obtain(u′′n(t),w)H + a(un(t),w) = (B(u ′ n(t)),w)H−g(∂tun(L, t))w(L) (40)for a.e. t > 0, which means that, w being arbitrary,u′′n(t) +Aun(t) = B(u ′ n(t))− g(∂tun(L, t))δL in V ′(41) for a.e. t >
Q13. What is the energy functional E defined on X?
The authors introduce the energy functional E defined on X byE(u,v) , 1 2[ ‖v‖2H + a(u,u) ] , (12)which is, in the context of abstract wave equations, the natural “mechanical” energy.
Q14. What is the identity of the rectangle?
Qτ the rectangle Ω× (0, τ), the authors have the following identity:Γuρ,µ(t) ∣∣∣∣τ 0 = −1 2 ∫∫ Qτ [ ∂tu ∂xu ]> [ ρ′ µ′ µ′ ρ′ ] [ ∂tu ∂xu ] + 12[ ρ(L)(1 + α2)−