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HAL Id: hal-02164050

https://hal.inria.fr/hal-02164050

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�

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Proofs for ”Discretization of Homogeneous Systems Using Euler Method with a State-Dependent Step”

Denis Efimov

a,b

, Andrey Polyakov

a,b

, Alexander Aleksandrov

c

aInria, 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

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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 hVνµ(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 matrix2∂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

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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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