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Preprint submitted on 24 Jun 2019
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Proofs for ”Discretization of Homogeneous Systems Using Euler Method with a State-Dependent Step”
Denis Efimov, Andrey Polyakov, Alexander Aleksandrov
To cite this version:
Denis Efimov, Andrey Polyakov, Alexander Aleksandrov. Proofs for ”Discretization of Homogeneous
Systems Using Euler Method with a State-Dependent Step”. 2019. �hal-02164050�
Proofs for ”Discretization of Homogeneous Systems Using Euler Method with a State-Dependent Step”
Denis Efimov
a,b, Andrey Polyakov
a,b, Alexander Aleksandrov
caInria, Univ. Lille, CNRS, UMR 9189 - CRIStAL, F-59000 Lille, France
bITMO University, 49 av. Kronverkskiy, 197101 Saint Petersburg, Russia
cSaint Petersburg State University, 7-9 Universitetskaya nab., 199034 Saint Petersburg, Russia
Abstract
This note contains some proofs for the paper ”Discretization of Homogeneous Systems Using Euler Method with a State- Dependent Step” of the same authors.
Key words: Homogeneous systems; Discretization; Euler method.
1 Proof of Theorem 9
Let us take a discretization step h > 0 and consider the behavior of ar–homogeneous Lyapunov functionV (satisfying (7)) on the sequence generated by (2). For this purpose definexi= Λr(kxikr)yi for someyi ∈Sr(1):
V(xi+1)−V(xi) =V(xi+kxik−νr hf(xi))−V(xi)
=kxikµr[V(yi+hf(yi))−V(yi)].
SinceV is twice continuously differentiable, then by the Taylor expansion theorem with Lagrange remainder [1]
there isθ∈(0,1) such that
V(yi+hf(yi)) =V(yi) + ∂V(ξ)
∂ξ ξ=y
i
hf(yi)
+h2
2 f>(yi) ∂2V(ξ)
∂ξ2 ξ=y
i+θhf(yi)
f(yi),
? This work was partially supported by ANR 18 CE40-0008 (Project DIGITSLID), the Government of Russian Federa- tion (Grant 08-08) and Saint Petersburg State University (Project Id 37569826).
Email addresses: [email protected](Denis Efimov),[email protected](Andrey Polyakov), [email protected](Alexander Aleksandrov).
then
V(xi+1)−V(xi) =kxikµr ∂V(ξ)
∂ξ ξ=y
i
hf(yi)
+kxikµrh2
2 f>(yi) ∂2V(ξ)
∂ξ2
ξ=yi+θhf(yi)
f(yi).
Note that from (7),
∂V(ξ)
∂ξ ξ=y
i
hf(yi)≤ −ha,
and there isv∈(0,+∞) such that
sup
y∈Sr(1)
sup
θ∈(0,1)
f>(y) ∂2V(ξ)
∂ξ2
ξ=y+θf(y)
f(y)≤v.
Therefore,
V(xi+1)−V(xi)≤hkxikµr h
2v−a
and for allh∈(0, h0] withh0= min{1,av}we obtain:
V(xi+1)−V(xi)≤ −a
2hkxikµr ≤ −αhV(xi), whereα= 2ca
2 and the property given in (7) was used on the last step. Assume that actuallyh0= min{1,av,2ca2},
Preprint submitted to Automatica 24 June 2019
then
V(xi+1)≤(1−αh)V(xi)
for all i= 0,1, . . . and the values ofV(xi) are monoto- nously decreasing, hence the sequence{xi}∞i=0is conver- ging (the property (b) is substantiated). Global boun- dedness of {xi}∞i=0 (i.e. the property (a)) follows from the properties of the functionV forγ= (c−11 c2)1/µ. Let us evaluate the rate of convergence for different signs of the degree of homogeneityν of (1) stated in the pro- perty (c). Letν= 0, then triviallyti+1=ti+hand an exponential rate of convergence is recovered. From (2) we have:
ti+1−ti= h
kxikνr ≤ h Vνµ(xi)
(
cν/µ1 ν <0 cν/µ2 ν >0. As it has been proven above
V(xi)≤(1−αh)iV(x0)
for alli= 0,1, . . ., then forν <0 and for anyk≥0 an estimate can be derived:
tk+1≤
k
X
i=0
cν/µ1 h
Vνµ(xi) ≤ cν/µ1 h Vµν(x0)
k
X
i=0
(1−αh)−iνµ.
Since|1−αh|<1 for the selectedh0, we get
t+∞≤ cν/µ1 h Vµν(x0)
+∞
X
i=0
(1−αh)−iνµ ≤ cν/µ1 hV−νµ(x0)
1−(1−αh)−µν <+∞
for allx06= 0, which implies a finite-time convergence to the origin of any sequence{xi}∞i=0in (2) but fori→+∞.
Forν >0 we are interesting in the time of convergence from an infinite initial condition toBr(1), then it is ne- cessary to repeat all above arguments in the inverse time, which leads to exactly the same estimate of such a time.
2 Proof of Proposition 11
Let xi ∈ Sr(1) be an arbitrary fixed vector, then the equation (3) can be rewritten as
∆ =hz(xi+ ∆)
for ∆ = xi+1−xi, where z: Rn →Rn is a continuous function defined as z(x) = kxk1ν
rf(x) withz(0) = 0. In- deed,z(Λr(λ)x) = Λr(λ)z(x) for anyλ >0 andx∈Rn, i.e.z is r–homogeneous vector field of degree 0 that is also continuous on Sr(1). Denote B(1) = {x ∈ Rn : kxk ≤1}as the unit ball inRn, if
h < inf
xi∈Sr(1),∆∈B(1)kz(xi+ ∆)k−1
then the functionhz(xi+·) : B(1)→B(1) is continu- ous on the convex compact setB(1). Hence, using the Brouwer fixed-point theorem [2] we conclude that the last equation has a solution with respect to ∆∈B(1) for anyxi∈Sr(1). The conclusion for anyxi∈Rn follows from Corollary 6 or Proposition 5.
3 Proof of Theorem 12
Definingxi+1= Λr(kxi+1kr)yi+1 withyi+1∈Sr(1), we obtain:
V(xi+1)−V(xi) =V(xi+1)−V(xi+1− kxi+1k−νr hf(xi+1))
=kxi+1kµr[V(yi+1)−V(yi+1−hf(yi+1))]
=kxi+1kµr ∂V(ξ)
∂ξ ξ=y
i+1
hf(yi+1)
−kxi+1kµrh2
2 f>(yi+1) ∂2V(ξ)
∂ξ2 ξ=y
i+1−θhf(yi+1)
f(yi+1)
with application of the Taylor expansion theorem with Lagrange remainder [1] on the last step. Next, similarly
∂V(ξ)
∂ξ ξ=y
i+1
hf(yi+1)≤ −ha
from (7), and if the matrix∂2∂ξV(ξ)2 is nonnegative definite for allξ∈Rn, then
V(xi+1)−V(xi)≤ −ahkxi+1kµr
for anyh >0. If this is not the case, then there is w∈ (0,+∞) such that
sup
y∈Sr(1)
sup
θ∈(0,1)
f>(y) ∂2V(ξ)
∂ξ2
ξ=y−θf(y)
f(y)≤w,
which results in
V(xi+1)−V(xi)≤hkxi+1kµr h
2w−a
and for allh∈(0, h0] withh0= min{1,wa}we obtain:
V(xi+1)−V(xi)≤ −a
2hkxi+1kµr ≤ −αhV(xi+1), where againα= 2ca
2 and the property given in (7) was used on the last step. Finally,
V(xi+1)≤(1 +αh)−1V(xi)
for alli= 0,1, . . . and the values ofV(xi) are monoto- nously decreasing, hence the sequence{xi}∞i=0is conver- ging (the property (b) is substantiated). Global boun- dedness of{xi}∞i=0 (i.e. the property (a)) follows from the properties of the functionV forγ= (c−11 c2)1/µ.
2
The property (c) can be proven applying the same argu- ments as in the proof of Theorem 9.
References
[1] M. Abramowitz and I. A. Stegun, editors. Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. Dover, New York, 9th edition, 1972.
[2] D. Leborgne.Calcul diff´erentiel et g´eom´etrie. Puf, Paris, 1982.
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