In [1], there was a typographical error in the entries of the off-diagonal elements of the matrix A(t), starting on the line before Equation (15). The purpose of this current note is to correct this mistake and propose a direction for future work. Choose W 1 , W 2 > 0 such that W 2 = W 1 e − h I * , and rewriting this equation with some little rearrangement,V(U(t)) now readṡwhere V(t, h) = col(U 1 (t),U 1 (t −h)), and
A(t) =I * +S * −S * +(U 1 (t)+U 2 (t)) −S * +(U 1 (t)+U 2 (t)) I * +S * .
Then, the determinant of A(t) is positive if and only if:=[(I * +2S * ) 2 −(−S * +(U 1 (t)+U 2 (t))) 2 ] > 0.That is, (I * +U 1 (t)+U 2 (t))(I * +2S * −(U 1 (t)+U 2 (t))) > 0, with solution satisfying the following inequality:Since is greater than zero, the matrix A is positive definite. For any > 0, considerOne can also see that direct computation of eigenvalues, say 1,2 of A(t) gives:1,2 (t) = I * +S * ±|S * −(U 1 (t)+U 2 (t))|. Since in 1 U 1 (t)+U 2 (t) S * , the minimum between the eigenvalues is min (t) = I * +U 1 (t)+U 2 (t), which is positive if U 1 (t)+ U 2 (t)<2S * . Hence, for any > 0, 1, ={(U 1 ,U 2 ) ∈ 1 : U 1 (t)+U 2 (t) > −2S * + }. It results that min (t) . Thus, = , provided that if (U 1 (0),U 2 (0)) ∈ 1, , then (U 1 (t),U 2 (t)) ∈ 1, for all t > 0. In future studies of similar problems, it would be interesting to prove global stability of the interior equilibrium by showing that solutions starting in 1 − 1, ultimately will enter 1, and remain there.