http://www.BillHowell.ca/Neural Nets/Paper reviews/170909 Journal paper peer review - mathematics only.ndf
www.BillHowell.ca 26Aug2017 initial, 09Sep2017 final
Bill@BillHowell.ca
View this file in a text editor with UNICODE characters (most modern text editors have this), constant width font (eg Courier 10), tabs of 3 spaces each, and ensuing that long text lines "wrap" to a new line.
I use the text editor "kwrite" in Linux, which is one of two or three that I have found to be excellent (there are many, many others!).
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Table of contents (without page numbers)
>>>>> Objectives of these step-by-step math checks
>>>>> "Flat-line" UNICODE nomenclature :
>>>>> Symbols
>>>>> Background
>>>>> PROOF OF LEMMA 1
>>>>> Sufficiency :
>>>>> check L1.S.1 of (p13L53) :
>>>>> check L1.S.2 of (4b), that is second part of (4) :
>>>>> Necessity
>>>>> check L1.N.1 of (21) :
>>>>> check L1.N.2 of (23) - first part : general form, nested product series :
>>>>> check L1.N.3 of (22) :
>>>>> check L1.N.4 of (23) :
>>>>> Preliminaries to Theorem 3
>>>>> PROOF OF THEOREM 3
>>>>> check T3.1 of (p6L17) :
>>>>> check T3.2 on inequalities (p14L37)(p14L40)(p14L42)(p14L44):
>>>>> check T3.3 of (p14L49) :
>>>>> check T3.4 of (27) :
>>>>> check T3.5 of (28) :
>>>>> check T3.6 of Remark 1
generated by :
$ cd "/media/bill/HOWELL_BASE/Neural Nets/My Reviews"
$ grep ">>>>> " "filename"
**************************
>>>>> Objectives of these step-by-step math checks
This text document is a record of this reviewer's step-by-step check over parts of the paper. It is too lengthy for a read-through by the authors, but if they want to scrutinize specific parts of my checks, there are provided in detail. My [style,nomenclature] is distracting, but quick and easy to use for a review.
As a reviewer, I find that a step-by-step re-typing of a part of the paper as I have done below forces me to :
- pay attention to details that I might otherwise skim over,
- learn & remember the author's detailed notations.
This is too time intensive to apply to the full paper, but by doing so over part of the authors' work, it gives me far grater confidence in the rest of the paper, which is read, but not analysed step-by-step. It also gives the authors a better idea of the weaknesses of the reviewer!
I have ONLY checked the derivations for Lemma 1 and Theorem 3.
I did NOT check Conclusions [1,2], Lemma 3, Theorem 4, Corrolaries [1,2], nor the "Comparison with existing results".
I did NOT do a detailed analysis of the [consistency, coherence] of the various componeents of the paper.
Reviewer names and the paper title have been removed from the text file, and the obscure posting of it on my zero-traffic website should provide the necessary privacy (I hope). If the authors want me to remove it after they have a chance to look at it, simply email me (my email address is in the file).
*****************************
>>>>> "Flat-line" UNICODE nomenclature :
A variable [scalar, vector, matrix]
A(i,j) (i,j)th component of A
A(i,j) (i,j)th component of A, notation just for simple indices
A_T transpose of A
|A| absolute value of matrix A (each element)
||A|| [Euclidean, spectral] norm of A ||A||2
A_dot time derivative of A, same as d[dt: A]
A_til A tilde over
A_ovr A overlined
A_und A underlined
A_hat A with ^ above it
A_sup(x) A superscript x - to avoid confuxion with A^x
A_sub(x) A subscript x - as distinct from A(x)
A^x A to the power of x
t_minus t approached from t < t_minus
t_plus t approached from t_plus < t
Indexing :
yi y(i), simple concatenation (might be confused with y*i)
y(i+1) array type notation
̄
Un-numbered [equations, inequalities] are denoted by their location, enclosed by parenthesis, eg :
(p5L42) y(t) ≤ ρ*sup[s∈[t−T,t]: y(s)]
means the above equation on page 5, Line 42.
Reviewer equations are denoted by some [section, subsection, whatever] and index, enclosed by parenthesis, eg :
(N.v) z(t**) ≤ d/c*z(t*)*e^(-θ*ln(c)/T) , ∀t ≥ t0
Calculus :
d[dt: x] total derivative of x with respect to t
dp[dt: x] partial derivative of x with respect to t
∫[ds: f] indefinite integral of f with respect to s
∫[ds, a to b: f] definite integral of f with respect to s, from a to b
Π[tt_plus : (r(t,t) - r(s,t))/(t - s)) }
D_minus D_minus(t) = lim{s->t_minus: (r(t,t) - r(s,t))/(t - s)) }
Various notations may be combined...
******* denotes start/end of topics & sub-topics
+-----+ denotes sub-steps in a [proof, development]..
denote [start,end] of checks on specific steps by the reviewer (me) :
>>>>> check :
<<<<< end check.
>> start of short reviewer comments
<< end of short reviewer comments
***************************
>>>>> Symbols
A(t) = diag (a1(t), · · · ,an(t)) with ai(t) > 0, i=1,···, n
B(t) = (b_ij)_sub_n×n denotes the delayed connection weight
c ∈ (0, 1) constant (p4L48)
d ≥ 1 constant (p4L54)
ε ≥ 0 constant (p4L37)
f = (f1,···,fn) ∈ C(Rn,Rn) is the activation function
γk positive costant, Lemma 2 p5L27
J = [t0,∞)
k ∈ N0
L = diag(l1,···,ln)
M ≥ 0 constant
N = {1,2···},
N0 = N ∪ {0},
PC(J,_ovr,∆) = {φ : J_ovr → ∆ is continuous everywhere except at finite
... number of points t, at which φ(t+), φ(t−) exist and φ(t+) = φ(t)}.
... assume "PC" stands for "Pulse Condition" or something like that
φ(s) ∈ PC([-τ_ovr,0],R^n) is the initial condition
R(l) I, l ≥ 1, is impulsive matrix
t ≥ t0 , denotes the reset rate of neurons
t∗∗ = t∗ + l*T + θ
{tk}_sup(∞)_sub(k=1) is the impulsive time sequence
T is the length of a time sub-interval within t* -> t**
such that T(1) ≤ t(k+1) − t(k) ≤ T(2) with T(2) ≥ T(1) > 0
τ(t) is the time-varying delay satisfying τ(t) ≤ τ_ovr with τ_ovr > 0,
θ θ∈[0,T) (p14L5), and as ε≥0 (p4L37), this implies that θ≥0?
x = (x1,···,xn) T ∈ R^n is the state vector of neurons
Lemma 1 :
from S(iii) in first part of Lemma 1 proof p14L3, and let T≥T_hat(c) :
S(iii) T_hat(c) = −1/ε*ln(c/M) >> so c = M*e^(ε*T_hat(c)) <<
T ≥ T_hat(c), t ∈ J, d = M, d ≥ 1,
Necessity part :
M = d/c
ε = ln(c)/T
t∗∗ = t∗ + l*T + θ, where θ ∈ [0, T )
*****************
>>>>> Background
>> p4L42 Reviewer assumes that "PC" expands to "Pulse Condition" (impulse), which by the expression given (from Definition 1), φ(s) is the maximum (supremum) function of time s within the interval τ_ovr preceding t0.
<<
In this paper, we consider the time-varying impulsive neural networks with time delays as follows :
(1) d[dt: x(t)] = -A(t)*x(t) + B(t)*f(x(t-τ(t))), t ∈ [t(k),t(k+1)]
x(t(k+1)) = R(k+1)*x(t_minus(k+1))
x(t0+s) = phi(s), s ∈ [-τ_ovr,0]
where :
see list of symbol defs in "NCAA-D-17-01191 m Liu, etal - Symbols.txt" ...
...
Definition 1 The impulsive neural network (1) is globally exponentially stable if there exist positive constants M and ε such that
(p5L21) ||x(t)|| ≤ M*||φ||_sub_ovr_τ * e^(−ε*(t−t0)), t ≥ t0
where ||φ||_sub_ovr_τ̄ = sup{s∈[−τ_ovr,0]: ||φ(s)|| }
Let μ(t) ∈ PC(J,R), γi > 0, i ∈ N0 (so z(t) ≥ 0)
(2) z(t) = Π[i=0 to k: γi*e^(∫[ds, t0 to t: μ(s)])], t ∈ [tk,t(k+1))
Definition 2 Function z(t) is a uniformly exponentially convergent function if there exist two constants M ≥ 1 and ε > 0 such that
(p2L38) z(t∗∗) ≤ M*z(t∗)*e^(−ε*(t∗∗ − t∗), ∀t∗∗ ≥ t∗ ≥ t0 .
Obviously, if z(t) is uniformly exponentially convergent, z(t) could converge to zero with convergent rate ε. As we know, z(t) is state transition of one-dimensional linear impulsive system
(3) y_dot(t) = μ(t)*y(t), t ∈ [tk,t(k+1) )
y(t(k+1)) = γ(k+1)*y(t_minus(k+1)), k ∈ N
Therefore, the uniformly exponential convergence of z(t) is equivalent to the exponential stability of the system (3).
Lemma 1 Function z(t) is a UECF if and only if, for any given c ∈ (0, 1), there exists T_hat(c) > 0 such that for any finite T ≥ T_hat(c) and t ∈ J,
(4) Π[t> Note : θ is not defined at the point in the paper where Eq (4) is declared. <<
****************
>>>>> PROOF OF LEMMA 1
+-----+
>>>>> Sufficiency :
If z(t) is a UECF, we have (>> see also Eq (2) p4L32 <<)
(p13L53) z(t+θ) = z(t)* Π[t>>>> check L1.S.1 of (p13L53) :
Note : reviewer equation numbers are of the form (L1.S.1.---)
Definition 1 - stability condition for impulsive neural network :
(p5L21) ||x(t)|| ≤ M*||φ||_sub_ovr_τ * e^(−ε*(t−t0)), t ≥ t0
Definition of z(t) :
(2) z(t) = Π[i=0 to k: γi*e^(∫[ds, t0 to t: μ(s)])], t ∈ [tk,t(k+1))
noting that for i=0, t=t0; while for 0*> OK - (L1.S.1.i) is the same as (p13L53). Note that I have not used Eqs (p14L3b,c) <<
So for expression for z(t+θ) , and using and Definition 2 Eq p4L38 :
(p13L53)
z(t+θ)* Π[t> OK, (L1.S.1.ii) is the same as authors' [equation,inequalities] (p14L3a) <<
<<<<< end check.
+---+
Therefore, we can choose
(p14L3a) T_hat(c) = −1/ε*ln(c/M) >> so c = M*e^(ε*T_hat(c)) <<
(p14L3a) d = M, d ≥ 1
>>>>> check L1.S.2 of (4b), that is second part of (4) :
Note : reviewer equation numbers are of the form (L1.S.2.---)
sub into (ii), using c from (iii), and let T≥T_hat(c)=θ
Π[t> OK, same as first part of (4) <<
sub into (ii) for d=M, d ≥ 1 :
Π[t> OK, (L1.S.2.v) is the same as second part of (4) <<
<<<<< end check.
+-----+
>>>>> Necessity
+---+
For any t∗∗ ≥ t∗ ≥ t0, there must exist some integer l ≥ 0
such that t∗∗ = t∗ + l*T + θ, where θ ∈ [0, T ). Then, we have :
(21) z(t**) = z(t*)* Π[t*>>>> check L1.N.1 of (21) :
Note : reviewer equation numbers are of the form (L1.N.1---)
from :
(2) z(t) = Π[i=0 to k : γi*e^(∫[ds, t0 to t : μ(s)])],
t ∈ [tk,t(k+1))
using this for t*,t**, where t∗∗ ≥ t∗ ≥ t0, and t,t** ∈ [tk,t(k+1))
and noting that for i=0, t=t*; while for 0**> OK - same as (21). Notice that the product series and integral range are tied - allowing easy separation of terms. <<
<<<<< end check.
+---+
When l ≥ 1, we deduce from (21) that
(23) z(t**) = z(t*)
* Π[j=0 to l-1 :
{ Π[t*+j*T>>>> check L1.N.2 of (23) - first part : general form, nested product series :
Note : reviewer equation numbers are of the form (L1.N.2.---)
General expression from expansion of (21) to (23) first line :
(21) z(t**) = z(t*)* Π[t*> OK - (L1.N.2.ii) is the same as the first line of authors' (23) <<
<<<<< end of check.
+---+
When l = 0, it follows from (21) that :
(22) z(t**,l=0) =z(t*)* Π[t*>>>> check L1.N.3 of (22) :
Note : reviewer equation numbers are of the form (L1.N.3.[i,ii,...])
but for l=0, t*+l*T = t*, the "last bit" range t* t*> so c = M*e^(ε*T_hat(c)) <<
T ≥ T_hat(c), t ∈ J, d = M, d ≥ 1,
But for the "Necessity" proof, the equivalent relations are :
(p14L21a,b) M = d/c, ε = ln(c)/T
It is OK to redefine things, for sure, but this should be clearly stated in the text, and it certainly confused me and may impact other readers as well. As the "Sufficiency" definitions (p14L3b,c) are not required in the Sufficiency proof, perhaps these are "left-over" from earlier stages of the authors' work, and that sentence can be removed?
sub for [M,ε] into (L1.N.3.iv) using (p14L21a,b), and noting that for l=0 t*+θ = t**, so z(t*+θ) = z(t**) :
z(t*+θ) ≤ M *z(t*)*e^(-θ*ε) , ∀t ≥ t0
(L1.N.3.v)
z(t**) ≤ d/c*z(t*)*e^(-θ*ln(c)/T) , ∀t ≥ t0
To substitute and inequality for z(t*), go back to :
(2) z(t) = Π[i=0 to k : γi*e^(∫[ds, t0 to t : μ(s)])],
t ∈ [tk,t(k+1))
For t=t*, this becomes :
(L1.N.3.vi)
z(t*) = Π[t0> p14L11 Proof of Lemma 1, Necessity - Equation (22)
My expressions are (see the link in the Section "C6. MATH CHECKS" below) :
(L1.N.3.v) z(t**) ≤ d/c*z(t*)*e^(-θ*ln(c)/T) , ∀t ≥ t0
(L1.N.3.vii) z(t*) ≤ c
(L1.N.3.viii) z(t**) ≤ d *e^(-θ*ln(c)/T) , ∀t ≥ t0
(L1.N.3.ix) z(t**) ≤ d *e^(-(t** - t*)*ln(c)/T) , ∀t ≥ t0
These are NOT the same as the authors' Eq (22) inequalities!! Furthermore I see no justification for the authors' use of (t** - θ - t*) = t0 in the first inequality. Note that :
e^(ln(c)/T*(t** - θ - t*)) = e^(-θ*ln(c)/T)*e^(ln(c)/T*(t** - t*))
so the third (last) expression of (22) is inconsistent with the second, as the term z(t*) has been set = 1, which is not justified. (Perhaps that term z(t*) has been accidentally dropped?) It seems that the authors should use EITHER (L1.N.3.vii) OR (L1.N.3.viii) inequality expressions, but not have a combination of both! Also, I have incorrectly used the constant c for (t* - t0) interval, and the authors may have done the same.
<<
Unimportant note : Equation (22) can more easily be expressed starting with :
(L1.N.3.v) z(t**) ≤ d/c*z(t*)*e^(-θ*ln(c)/T) , ∀t ≥ t0
becomes z(t**) ≤ d *z(t*)*c^(-1 - θ/T),
(one=1 in (), "small L" = l)
using z(t*) ≤ c
z(t**) ≤ d *c^(-θ/T) , ∀t ≥ t0
However, the exponential form will be more handy later in the paper...
<<<<< end check.
+---+
When l ≥ 1, we deduce from (21) (confirmed by reviewer in (N.ii)) that
(23) z(t**)
= z(t*)
* Π[j=0 to l-1 :
{ Π[t+j*T>>>> check L1.N.4 of (23) :
Note : reviewer equation numbers are of the form (L1.N.4.---)
Note : The nested product series allow the authors to address portions of the series at a time.
From authors' [equation,inequalities] (p13L53),(p14L3a), confirmed by reviewer :
(p13L53) z(t+θ) = z(t) *Π[t z(t**) :
(L1.N.4.i)
z(t**)
= z(t*)
* Π[j=0 to l-1 :
{ Π[t*+j*T> OK - this is the same as the first inequality expression of (23) <<
putting (N.xiv) in exponential format for use in later Theorems etc, and replacing l with (t** - θ - t*)/T :
z(t**) ≤ d*z(t*) *e^((t** - θ - t*)/T*ln(c))
(L1.N.4.vi)
z(t**) ≤ d*z(t*) *e^(-θ/T*ln(c))*e^((t** - t*)/T*ln(c))
>> OK - (L1.N.4.vi) is the same as the second inequality expression of (23) <<
Going further, and sub for z(t*) from (N.x), (N.xiv) becomes :
z(t**) ≤ d*b*c^l
If I (incorrectly?) assume b=c, :
z(t**) ≤ d*c^(1+ (t** - θ - t*)/T)
... not used by authors
>> p14L19 Proof of Lemma 1, Necessity - Equation (23)
My derived expressions agree with the authors Eq (23). However, I had to assume that the SAME constant "c" applies to both intervals : (t* - t0), and (t* + l*T), which is possibly not correct (but perhaps this is allowable assuming a "somewhat flexible" "c").
<<
>> p14L5 Proof of Lemma 1 - Necessary and Sufficient conditions (comment)
While I am comfortable with the authors' "Sufficiency" proof, I am not entirely convinced of their proof of "Necessity". It must be proved (eg proof of necessity by negation is one way) that if conditions Eq (4) don't apply, then z(t) cannot be a UECF. For example :
- it would have to be shown that for c ≥ 1, z(t) is not a UECF or something like that. While one might think that by having l->∞ and T->0, it would be easy to show that, is this really shown in the proof?
- perhaps a different approach than the authors would result in different proven stability criteria formulations, that do not always fit within the authors' approach. How do I know if that cannot be the case?
<<
<<<<< end check.
**************************
>>>>> Preliminaries to Theorem 3
(no reviewer step-by-step checks on these items)
Definition 3 If z(t) is a UECF, then, the set of
(p5L8) Ξz = {T > 0: sup{t∈J; {Π[t 0, we have χz(T) ≤ γ_ovr^κ*e^max{0,μmax*T},
where γ_ovr = max{1, max{i∈N: γi}}, μmax = max{t∈J: μ(t)}, κ is the smallest positive integer such that T − κ*δ_til ≤ 0 with
δ_til = min(k∈N0: t(k+1) − tk). Specially, if μ(t) = μ > 0, γk ≤ 1, t ∈ J, k ∈ N0 , then χz(T) ≤ e^μ*T.
The following two lemmas are critical to deduce the stability results of the neural network (1). The proof of Lemma 2 is given in Appendix. Lemma 3 is a special form of [51, Lemma 1].
Lemma 2 Let y(t) and
(p5L27) w(t) ∈ PC[J,R_plus) satisfy that :
(i). D + y(t) ≤ μ(t)y(t), whenever y(t) ≥ ψ(w(t)),
(ii). y(t(k+1)) ≤ γ(k+1)*y t −
k+1 , k ∈ N ,
where ψ(·) is a K-function, function μ(t) and positive constant γk , k ∈ N , satisfy that z(t) is a UECF with z(t) being defined by (2). Then, for any given constant T > 0 and any t ≥ T + t0, one could obtain :
(5) y(t) ≤ max{s∈[0,T]: y(t − T)
*Π[t-T 0 be a constant. Let a piecewise continuous function
y : [t0−T, +∞) → [0, +∞) admit a sequence of real numbers tk and positive constants δ_til and δ_ovr such that t(k+1) − tk ∈ [δ_til,δ_ovr] for all k ∈ N0, and y_ovr(t) exists and is finite for each k ∈ N. Assume that there is a constant ρ ∈ (0,1) such that
(p5L42) y(t) ≤ ρ*sup[s∈[t−T,t]: y(s)]
holds for all t ≥ t0 . Then
(p5L46) y(t) ≤ sup{s∈[t0−T,t0]: e^(ln(ρ)*(t−t0)/T), ∀t ≥ t0
**************************
>>>>> PROOF OF THEOREM 3
Theorem 3 Assume that there exist a piecewise continuous function V(t,x),
t ≥ t0 − τ_ovr, x ∈ R^n, piecewise continuous function μ(t), K∞ function q, positive constants c1, c2, p, T ∈ Ξz , ρ ∈ (0,1), where z(t) is a UECF defined by (2), such that :
(p6L8.i) c1*||x||^p ≤ V(t,x(tk)) ≤ c2*||x||^p ;
(p6L9.ii) V(tk,x(tk)) ≤ γk*V(tk_minus,x(tk_minus), k ∈ N ;
(p6L10.iii) q(ρ*s/χz(T)) ≥ s, s ≥ 0;
(p6L12.iv) D_plus(V(t,x(t)) ≤ μ(t)*V(t,x(t)),
whenever V(t+s,x(t+s)) ≤ q(V(t,x(t))), s ∈ [-τ_ovr,0], t ∈ J.
Then, the impulsive neural network (1) is globally exponentially stable.
+-----+
For convenience, we can choose q(s) as a linear function. Namely, q(s) = q*s. Then, Condition (iii) in Theorem 3 could be rewritten as
(p6L17a) q(s) = q*s - assumption of simple linear
(p6L17b) q ≥ χz(T)/ρ
(p6L18) q > χz(T) - as ρ ∈ (0,1)
(p14L37) y(t) ≤ max{ y(t − T)*Π[t-T χz(T).
>>>>> check T3.1 of (p6L17) :
(easy by inspection, but here it is anyways)
Note : reviewer equation numbers are of the form are (T3.1.---)
(p6L8.iii) q(ρ*s/χz(T)) ≥ s, s ≥ 0;
sub q(s) = q*s
q*(ρ*s/χz(T)) ≥ s, s ≥ 0;
q ≥ s/(ρ*s/χz(T)), s ≥ 0;
(p6L17) q ≥ χz(T)/ρ
but as ρ ∈ (0,1)
(T3.1.i) q ≥ χz(T)
>> OK - (T3.1.i) is the same as (p6L18)<<
<<<<< end check T3.1.
+---+
Let
(p14L34a) y(t) = V(t,x(t)),
(p14L34b) w(t) = sup{s∈[−τ_ovr,0]: V(t+s, x(t+s)),
(p14L34c) ψ(w) = q^(-1)(w)
Then, based on the conditions of Theorem 3 and Lemma 1, we obtain
(p14L37) y(t) ≤ max{ y(t − T)*Π[t-T>>>> check T3.2 on inequalities (p14L37)(p14L40)(p14L42)(p14L44):
Note : reviewer equation numbers are of the form are (T3.2.---)
from Lemma (2), keeping in mind s∈[0,T] :
(5) y(t) ≤ max{ y(t − T)*Π[t-T> OK - (T3.2.i) is the same as the first inequality (p14L37) <<
+---+
substitute :
(p14L46) η_z(t) = Π[t-T> OK, (T3.2.v) is the same as (p14L40).
But I am not entirely comfortable with :
- I used Lemma 2, not Lemma 1
eg (p14L34) "based on the conditions of Theorem 3 and Lemma 1,"
- The linear q assumption of (p6L17a) and (T3.2.iii)
would seem to greatly restrict the generality of the result!
- The notation of :
(p6L17a) q(s) = q*s - assumption of simple linear
(p6L10.iii) q(ρ*s/χz(T)) ≥ s, s ≥ 0;
as for me there is confusion as to whether the linear q = q*s (or q(w(s)=q*w(s)) assumption is already implied, whereas it is only mentioned in a later paragraph.
- In (p6L10.iii) q(ρ*s/χz(T)) ≥ s, s ≥ 0, whereas it is applied here when t−T≤s≤t, which might be negative, (p14L34b) has s∈[−τ_ovr,0] which is definitely negative, and later inequality (28) has −T_ovr≤s≤0 which is also definitely negative. (is (28) pertinent to this issue???????, probably not but I haven't looked closely)
<<
+---+
Going from
(p14L40) y(t) ≤ max{ y(t − T)*η_z(t), ρ*sup{t−T≤s≤t: w(s) }
Note that
(p14L27) s ∈ [t−T,t] for Lemma 2, but not always satisfied ...
(p4L7) τ(t) ≤ τ_ovr with τ_ovr > 0
(p14L34a) y(t) = V(t,x(t))
so
(T3.2.vi) y(t+s) = V(t+s,x(t+s))
(p14L34b) w(t) = sup{s∈[−τ_ovr,0]: V(t+s,x(t+s))}
(T3.2.vii) w(t) = sup{−τ_ovr≤s≤0: y(t+s)}
(p14L46b) T_ovr = T + τ_ovr
(T3.2.viii) w(t) = sup{T-T_ovr≤s≤0: y(t+s)}
so (p14L40) can be written as :
(T3.2.ix) y(t) ≤ max{ y(t − T)*η_z(t), ρ*sup{T-T_ovr≤s≤0: y(t+s) }
>> WRONG - (T3.2.ix) is the same as the first inequality (p14L42), EXCEPT that the lower limit is (T-T_ovr) rather than T_ovr. This would also affect (p14L44) <<
Going from
(p14L42) y(t) ≤ max{ y(t−T)*η_z(t), ρ*sup{T_ovr≤s≤0: y(t+s) }
But
(p14L27) s ∈ [t−T,t]
therefore y(t−T) ≤ sup{T_ovr≤s≤0: y(t+s) }
and the following is true (albeit less conservative)
(T3.2.x) y(t) ≤ max{η_z(t),ρ}*sup{T_ovr≤s≤0: y(t+s) }
>> OK - (T3.2.x) is the same as (p14L44), understanding that I used (p14L42) as the starting point, and NOT my (T3.2.ix).
<<<<< end check T3.2.
+-----+
p14L46 Because T ∈ Ξz , we know that there exists some ηz_max ∈ (0,1)
such that ηz(t) ≤ ηz_max, t ≥ t0+T. Denote ρ_ovr = max{ρ,ηz_max}, we have :
(p14L49) y(t) ≤ ρ_ovr*sup{-T_ovr≤s≤0: y(t+s)}
>>>>> check T3.3 of (p14L49) :
Note : reviewer equation numbers are of the form are (T3.3.---)
p5L8 Definition 3 ensures that ηz_max ∈ (0,1)
from
(p14L44) y(t) ≤ sup{T_ovr≤s≤0: y(t+s)}*max{ η_z(t),ρ }
sub
(p15L46) ρ_ovr = max{ρ,ηz_max}
(T3.3.i) y(t) ≤ ρ_ovr*sup{T_ovr≤s≤0: y(t+s) }
>> WRONG - (T3.3.i) is the same as (p14L49), EXCEPT that the lower limit is T_ovr rather than -T_ovr. This would also affect (27) <<
<<<<< end check T3.3.
+-----+
By Lemma 3, the above inequality implies that
(27) y(t) ≤ sup{-T_ovr≤s≤0: y(t*+s)}*e^(ln(ρ_ovr)*(t−t*)/T)
>>>>> check T3.4 of (27) :
Note : reviewer equation numbers are of the form are (T3.4.---)
From Preliminaries
(p3L41) J = [t0,∞)
For Lemma 2 :
(p14L27) s ∈ [t−T,t] for Lemma 2, different context
Lemma 3
(p5L46) y(t) ≤ sup{s∈[t0−T,t0]: e^(ln(ρ)*(t−t0)/T)}, ∀t ≥ t0
>> NOTE : There is an important difference between definitions of t for Lemma 3 and Theorem 3!! So they should be used only when necessary, and with caution! <<
For Theorem 3 :
(p6L5) t ≥ t0 − τ_ovr
(p6L6) T ∈ Ξz
For Theorem 3, condition (iv) :
(p6L12) s∈[t0−T,t0]
(p6L12) t ∈ J
Given for the proof of Theorem 3
(p14L46b) T_ovr = T + τ_ovr
In (p5L46), this implies :
(T3.4.i) {s∈[t0−T,t0]} => {t0−T≤s≤t0}
Approach 1 : Illegitimate substitutes of other inequalities
sub (p6L5) into (T3.4.i) Right Hand Side (RHS) :
{s∈[t0−T,t0]} => {t-τ_ovr-T ≤ s ≤ t-τ_ovr}
subtract t - τ_ovr from (T3.4.ii) (Illegitimate!!) :
(T3.4.ii) {s∈[t0−T,t0]} => {-T≤s≤0}
>> INCORRECT : I have -T in my (T3.4.ii), instead of -T_ovr in (27) p14L53.
<<
Approach 2 :
simply set t0 = 0
(T3.4.iii) {s∈[t0−T,t0]} => {−T≤s≤0}
>> INCORRECT - I have -T instead of T_ovr!!
Also, t0=0 is not stated by the authors, and this substitution potentially affects the generality of the result. I am missing a key relationship here. <<
I will proced with (T3.4.iii), ???? but using -T_ovr instead of -T and see what it yields.
Again - Lemma 3
(p5L46) y(t) ≤ sup{s∈[t0−T,t0]: e^(ln(ρ)*(t−t0)/T)}, ∀t ≥ t0
>> WRONG! : It seems to me that (p5L46) is missing a minus sign in the exponent, as [t,T]>t0 and therefore the existing expression tends RISES over time instead of attenuating.
<<
sub (T3.4.iii) and (p15L3) into modified (p5L46), ∀t ≥ t0
y(t) ≤ sup{−T_ovr≤s≤0: e^(-ln(ρ)*(t−t*-T)/T)}
y(t) ≤ sup{−T_ovr≤s≤0: e^(ln(ρ))*e^(ln(ρ)*(t−t*)/T)}
(T3.4.iv) y(t) ≤ ρ*sup{−T_ovr≤s≤0: e^(ln(ρ)*(t−t*)/T)}
Comparing :
(p14L49) y(t) ≤ ρ_ovr*sup{-T_ovr≤s≤0: y(t+s)}
(T3.4.iv) y(t) ≤ ρ *sup{−T_ovr≤s≤0: e^(ln(ρ)*(t−t*)/T)}
It seems difficult to reconcile these two conditions with p15L49 :
(27) y(t) ≤ sup{-T_ovr≤s≤0: y(t*+s)}*e^(ln(ρ_ovr)*(t−t*)/T)
unless I can show that :
sup{-T_ovr≤s≤0: y(t*+s)}*e^(ln(ρ_ovr)*(t−t*)/T)
≥ ρ_ovr*sup{-T_ovr≤s≤0: y(t+s)}
OR
≥ ρ *sup{−T_ovr≤s≤0: e^(ln(ρ)*(t−t*)/T)}
This does sound plausible, but in doing so, will this make the result overly conservative? Maybe that is less important than having a functional form that helps to prove Theorem 3 overall.
Attempt 1 : The easiest way to prove this might be to prove that :
(T3.4.v) sup{-T_ovr≤s≤0: y(t*+s)}*e^(ln(ρ_ovr)*(t−t*)/T)
≥ ρ_ovr*sup{-T_ovr≤s≤0: y(t+s)}
Dividing both sides by sup{-T_ovr≤s≤0: y(t*+s)} ≥ 0 under the assumption that y(t) is a POSITIVE one-dimensional linear impulsive system :
ρ_ovr ≤ e^(ln(ρ_ovr)*(t−t*)/T)
(T3.4.vi) ρ_ovr ≤ e^((t−t*)/T)^(ln(ρ_ovr))
I will use :
(p14L30a) t∗ < t
(p15L3) t∗ = t0 + T
with an upper bound t_max defined by the K∞ upper envelope :
(p14L26) t∗ = sup{s: y(s) < ψ(w(s))}
So, the expression e^((t−t*)/T) has bounds :
from e^((t*−t*)/T) to e^((t−(t0+T)/T)
or from 1 to e^((t−t0)/T)
If I simpy assume t0=0 the 2nd expression
e^((t−(t0+T)/T) ≈ e^((t−T)/T) = e^(t/T−1) ≥ 1
Revising the bounds of expression e^((t−t*)/T) :
from 1 to e^(t/T−1)
(T3.4.vii) e^((t−t*)/T) ≥ 1
Revisit
(T3.4.vi) ρ_ovr ≤ e^((t−t*)/T)^(ln(ρ_ovr))
take ln of both sides
ln(ρ_ovr) ≤ ln(ρ_ovr)* ln(e^((t−t*)/T))
1 ≤ ln(e^((t−t*)/T))
Taking exp of each side
(T3.4.vi) e ≤ e^((t−t*)/T)
such that from (T3.4.vii) and (T3.4.vi),
1 ≤ e ≤ e^((t−t*)/T)
using the least conservative criteria 1 ≤ e^((t−t*)/T) in
(T3.4.vi) ρ_ovr ≤ e^((t−t*)/T)^(ln(ρ_ovr))
ρ_ovr ≤ 1^(ln(ρ_ovr)) = 1
which we had already, but this way we've shown that (T3.4.v) applies :
(T3.4.v) sup{-T_ovr≤s≤0: y(t*+s)}*e^(ln(ρ_ovr)*(t−t*)/T)
≥ ρ_ovr*sup{-T_ovr≤s≤0: y(t+s)}
and we can rewrite :
(p14L49) y(t) ≤ ρ_ovr*sup{-T_ovr≤s≤0: y(t+s)}
as
(T3.4.v) y(t) ≤ sup{-T_ovr≤s≤0: y(t*+s)}*e^(ln(ρ_ovr)*(t−t*)/T)
>> OK - (T3.4.v) is the same as (27)
(27) y(t) ≤ sup{-T_ovr≤s≤0: y(t*+s)}*e^(ln(ρ_ovr)*(t−t*)/T)
<<<<< end check T3.4.
+-----+
p15L3 It follows from Condition (i) and inequality (27) that
(28) ||x(t)|| ≤ (c2/c1)^(1/p)*||x(t*)||_T_ovr
*e^(ln(ρ_ovr)/p*(t−t*)/T_ovr)
where
(p15L8) ||x(t*)||_T_ovr = sup{−T_ovr≤s≤0: ||x(t*+s)||}
This indicates that the neural network (1) is globally exponentially
stable.
>>>>> check T3.5 of (28) :
Start with :
Definition 1 The impulsive neural network (1) is globally exponentially stable if there exist positive constants M and ε such that
(p5L21) ||x(t)|| ≤ M*||φ||_sub_ovr_τ * e^(−ε*(t−t0)), t ≥ t0
where ||φ||_sub_ovr_τ̄ = sup{s∈[−τ_ovr,0]: ||φ(s)|| }
Theorem 3, :
(p6L8.i) c1*||x||^p ≤ V(t,x(tk)) ≤ c2*||x||^p ; = Condition 1
(p14L34a) y(t) = V(t,x(t)) Theorem 3 preliminary
(27) y(t) ≤ sup{-T_ovr≤s≤0: y(t*+s)}*e^(ln(ρ_ovr)*(t−t*)/T)
Here I ASSUME that ||x|| = ||x(tk)|| in Condition 1 (p6L8.i)
Sub (p14L34a) into (p6L8.i) :
(T3.5.i) c1*||x||^p
≤ y(t)
≤ c2*||x||^p
WARNING!! while [c1,||x||,p] ≥ 0, there is no guarantee in (3) that y(t) ≥ 0, nor that V(t,x(tk)) ≥ 0, in which case Condition 1 is immediately violated!! I will procede in spite of that, assuming the unstated condition that y(t)≥0, based on the requirements of Condition 1. This is equivalent to assuming that y(t) is a linear positive impulsive system (eg https://arxiv.org/pdf/1703.03950.pdf).
Immediate substitution of (27) into (p6L8.i) could violate the condition "≤ c2*||x||^p", so it is best to simply abandon the higher bound for now.
Sub (27) into (T3.5.i) :
(T3.5.ii) c1*||x||^p ≤ sup{-T_ovr≤s≤0: y(t*+s)}*e^(ln(ρ_ovr)*(t−t*)/T)
Dividing both sides by c1≥0 and taking the (1/p) root of (T3.5.i) :
||x|| ≤ (1/c1)^(1/p)
*sup{-T_ovr≤s≤0: [y(t*+s)]^(1/p)}
*[e^(ln(ρ_ovr)*(t−t*)/T)]^(1/p)
(T3.5.iii) ||x|| ≤ (1/c1)^(1/p)
*sup{-T_ovr≤s≤0: [y(t*+s)]^(1/p)}
*e^(ln(ρ_ovr)/p*(t−t*)/T)
Substitute
(p15L8) ||x(t*)||_T_ovr = sup{−T_ovr≤s≤0: ||x(t*+s)||}
To do this, the relation between x(t*) and y(t*) is required. Equations (1) and (3) obviously have strong similarities :
time-varying impulsive neural networks with time delays :
(1) d[dt: x(t)] = -A(t)*x(t) + B(t)*f(x(t-τ(t))), t ∈ [t(k),t(k+1)]
x(t(k+1)) = R(k+1)*x(t_minus(k+1))
x(t0+s) = phi(s), s ∈ [-τ_ovr,0]
(3) y_dot(t) = μ(t)*y(t), t ∈ [tk,t(k+1) )
y(t(k+1)) = γ(k+1)*y(t_minus(k+1)), k ∈ N
Given the term B(t)*f(x(t-τ(t))), this similarity is not exact. However, it appears that the authors have assumed that ||x(t*)|| = y(t*). Substituting that assumption into (T3.5.iii) :
(T3.5.iii) ||x|| ≤ (1 /c1)^(1/p)*||x(t*)||_T_ovr^(1/p)
*e^(ln(ρ_ovr)/p*(t−t*)/T)
Compare this to :
(28) ||x(t)|| ≤ (c2/c1)^(1/p)*||x(t*)||_T_ovr
*e^(ln(ρ_ovr)/p*(t−t*)/T_ovr)
>> INCORRECT - While (T3.5.iii) is similar to (28), there are differences :
1. constant expression : I see no justification for c2
(T3.5.iii) (1 /c1)^(1/p)
(28) (c2/c1)^(1/p)
2. exponent of ||x(t*)||_T_ovr : This appears to be an omission of (28)
(T3.5.iii) 1/p
(28) 1
3. Different subscripts [ρ,ρ_ovr] and [T,T_ovr]
Some discrepancies with these show up as noted above in previous parts of the proof as well. It is certainly more conservative to use the [ρ_ovr,T_ovr] forms, so that is not a problem for the theory.
<<
>> COMMENT : Proof of Theorem 3 that (1) is globally exponentially stable :
Compare the following 3 expressions :
...
Definition 1 The impulsive neural network (1) is globally exponentially stable if there exist positive constants M and ε such that
(p5L21) ||x(t)|| ≤ M*||φ||_sub_ovr_τ * e^(−ε*(t−t0)), t ≥ t0
where ||φ||_sub_ovr_τ̄ = sup{s∈[−τ_ovr,0]: ||φ(s)|| }
...
(28) ||x(t)|| ≤ (c2/c1)^(1/p)*||x(t*)||_T_ovr
*e^(ln(ρ_ovr)/p*(t−t*)/T_ovr)
...
Reviewer's own derivation :
(T3.5.iii) ||x|| ≤ (1 /c1)^(1/p)*||x(t*)||_T_ovr^(1/p)
*e^(ln(ρ_ovr)/p*(t−t*)/T)
...
Both (28) and (T3.5.iii) reduce easily to Definition 1 (p5L21), EXCEPT for the exponent (1/p) of ||x(t*)||_T_ovr^(1/p) in (T3.5.iii). The ommission of c2 in (T3.5.iii) is a minor point, but I didn't see justification for having it. If (T3.5.iii) has the correct exponent, then Theorem 3 is not strictly proven. For p "close to" 1, then the Theory is approximately correct.
<<
<<<<< end check T3.5.
+---+
Remark 1 - In Theorem 3, we do not restrict that γk ≤ 1 for all k ∈ N . This indicates that Theorem 3 permits some impulses have a destabilizing effect. Moreover, we only restrict that function μ(t) is piecewise continuous, rather than μ(t) > 0 or μ(t) < 0. Therefore, the time-derivate of Lyapunov function under Razumikhin condition could be negative definite, or positive definite, or negative definite during some time intervals and positive definite during other time intervals. Thus, the restrictive conditions in Theorem 3 are more permissive.
>>>>> check T3.6 of Remark 1
No special comments - this is clear.
<<<<< end check T3.6.
enddoc*