In this section, we consider the following finite-time nonautonomous linear differential equation with off-diagonal delays (see, e.g., [13] and the reference therein):
(2.1)x˙i(t)=aii(t)xi(t)+∑j=1,j≠idaij(t)xj(t-τij) for i=1,…,d, t∈[0, T],
where T is a given positive constant, aij:ℝ→ℝ, i,j=1,…,d, are continuous functions and τij>0 for i,j=1,…,d with i≠j. Define
(2.2)r:=max{τij:i,j=1,…,d, i≠j}.
Note that (2.1) is a special case of (1.2). More precisely, the right hand side of (2.1) equals f(t,xt), where f=(f1,…,fd):[0,T]×𝒞→ℝd is defined as follows:
(2.3)fi(t,φ):=aii(t)φ(0)+∑j=1,j≠idaij(t)φj(-τij).
Let S:[0,T]×𝒞→𝒞 denote the evolution operator of (2.1). From (2.3), we see that the function f is linear in the second argument. Therefore, the evolution operator S is also linear in the second argument. Our aim in this section is to provide a sufficient condition for the finite-time attractivity for the zero solution of (2.1) and thus for all solutions of (2.1), see Remark 1.2.

Before presenting the main result, we recall the notion of row diagonal dominance. We refer the reader to [14, Definition 7.10] for a discussion of this notion. System (2.1) is called row diagonally dominant if there exists a positive constant δ such that
(2.4)|aii(t)|≥∑j=1,j≠id|aij(t)|+δ for t∈[0,T].

Proof.
We divide the proof into two steps.

Step 1. We show that for φ∈𝒞 the inequality
(2.8)‖x(t,φ)‖∞≤e-(γ/2)(t-s)‖S(s,φ)‖γ,∞ ∀0≤s≤t≤T
holds. Suppose the opposite, that is, assume that there exists s∈[0,T] such that the set
(2.9)N:={t∈[s,T):‖x(t,φ)‖∞>e-(γ/2)(t-s)‖S(s,φ)‖γ,∞}
is not empty. Define tinf =inf {t:t∈𝒩}. By continuity of the map t↦∥x(t,φ)∥∞, we get that
(2.10)‖x(tinf,φ)‖∞=e-(γ/2)(tinf-s)‖S(s,φ)‖γ,∞,(2.11)‖x(t,φ)‖∞≤e-(γ/2)(t-s)‖S(s,φ)‖γ,∞ ∀t∈[s,tinf].
Now, we will show that
(2.12)‖x(tinf,φ)‖∞≥e-γr‖x(t,φ)‖∞ ∀t∈[tinf-r,tinf].
Indeed, we consider the following two cases: (i) t∈[s,tinf ]∩[tinf -r,tinf ] and (ii) t∈(-∞,s]∩[tinf -r,tinf ].

Case (i). If t∈[s,tinf ]∩[tinf -r,tinf ], then, according to (2.10) and (2.11), we obtain that
(2.13)‖x(tinf,φ)‖∞=e-(γ/2)(tinf-s)‖S(s,φ)‖γ,∞≥e-(γ/2)(tinf-s)e(γ/2)(t-s)‖x(t,φ)‖∞≥e-γr‖x(t,φ)‖∞,
which proves (2.12) in this case.

Case (ii). If t∈(-∞,s]∩[tinf -r,tinf ], then, according to (2.10) and the definition of the norm ∥·∥γ,∞, we obtain that
(2.14)‖x(tinf,φ)‖∞=e-(γ/2)(tinf-s)maxω∈[-r,0]eγω‖S(s,φ)(ω)‖∞=maxω∈[-r,0]e-(γ/2)(tinf-s-2ω)‖x(s+ω,φ)‖∞≥e-(γ/2)(tinf+s-2t)‖x(t,φ)‖∞≥e-γr‖x(t,φ)‖∞.
Hence, (2.12) is proved. To conclude the proof of this step, we estimate the norm ∥x(t,φ)∥∞ for all t in a neighborhood of tinf in order to show a contradiction to the assumption that the set 𝒩 is not empty. To this end, we define the following set:
(2.15)I:={i∈{1,…,d}:|xi(tinf,φ)|=‖x(tinf,φ)‖∞}.
The continuity of the functions t↦xi(t,φ) for i=1,…,d implies that there exists a neighborhood (tinf -ε,tinf +ε) for some ε>0 such that
(2.16)‖x(t,φ)‖∞=maxi∈I|xi(t,φ)| ∀t∈(tinf-ε,tinf+ε).
By virtue of (2.1), the derivative of the function t↦xi2(t,φ) is estimated as follows:
(2.17)12ddtxi2(t,φ)|t=tinf=xi(tinf,φ)[{∑j=1,j≠id}aii(tinf)xi(tinf,φ)xi(tinf,φ) +∑j=1,j≠idaij(tinf)xj(tinf-τij,φ)]≤aii(tinf)xi(tinf,φ)2 +|xi(tinf,φ)|∑j=1,j≠id|aij(tinf)|‖x(tinf-τij,φ)‖∞,
which together with (2.12) and the definition of I implies that for all i∈I(2.18)12ddtxi(t,φ)2|t=tinf≤xi(tinf,φ)2[aii(tinf)+eγr∑j=1,j≠id|aij(tinf)|].
Thus, from the row diagonal dominance (2.4) and bound (2.5), we derive that
(2.19)12ddtxi(t,φ)2|t=tinf≤[-δ+(eγr-1)M]xi2(tinf,φ).
Using (2.6), we obtain that
(2.20)12ddtxi2(t,φ)|t=tinf<-γ2xi2(tinf,φ),
which yields that there exists a neighborhood (tinf ,tinf +ε2) for an ε2>0 such that for i∈I(2.21)|xi(t,φ)|≤e-(γ/2)(t-tinf)|xi(tinf,φ)| ∀t∈(tinf,tinf+ε2).
Thus, for all t∈(tinf ,tinf +ε) with ε:=min {ε1,ε2}, we have
(2.22)‖x(t,φ)‖∞≤e-(γ/2)(t-tinf)‖x(tinf,φ)‖∞,
which together with (2.10) implies that
(2.23)‖x(t,φ)‖∞≤e-(γ/2)(t-s)‖S(s,φ)‖γ,∞.
Consequently, 𝒩∩(tinf ,tinf +ε)=∅, and this is a contradiction to the definition of 𝒩 and tinf . Thus, (2.8) is proved.

Step 2. Using (2.8) from Step 1, we show (2.7) by considering two cases: (i) t∈[s+r,T]∩[s,T] and (ii) t∈[s,s+r]∩[s,T].

Case (i). If t∈[s+r,T]∩[s,T], then by virtue of (2.8) we have
(2.24)‖S(t,φ)‖γ,∞=supω∈[-r,0]eγω‖x(t+ω,φ)‖∞≤supω∈[-r,0]eγωe-(γ/2)(t+ω-s)‖S(s,φ)‖γ,∞≤e-(γ/2)(t-s)‖S(s,φ)‖γ,∞,
which proves (2.7) in this case.

Case (ii). If t∈[s,s+r]∩[s,T], then we have
(2.25)‖S(t,φ)‖γ,∞=max{supω∈[-(t-s),0]eγω‖x(t+ω,φ)‖∞, =maxsupω∈[-r,-(t-s)]eγω‖x(t+ω,φ)‖∞}.

Hence, using (2.8), we obtain that
(2.26)‖S(t,φ)‖γ,∞≤max{supω∈[-(t-s),0]e(γ/2)ωe-(γ/2)(t-s)‖S(s,φ)‖γ,∞, ≤maxe-γ(t-s)‖S(s,φ)‖γ,∞{supω∈[-(t-s),0]}}.

Thus, (2.7) is proved, and the proof is complete.