c Allerton Press, Inc., 2007. ISSN 1066-369X, Russian Mathematics (Iz. VUZ), 2007, Vol. 51, No. 6, pp. 52–60. c T.L. Sabatulina, V.V. Malygina, 2007, published in Izvestiya Vysshikh Uchebnykh Zavedenii. Matematika, 2007, No. 6, pp. 55–63. Original Russian Text
Several Stability Tests for Linear Autonomous Differential Equations with Distributed Delay T. L. Sabatullina and V. V. Malygina* Perm State Technical University, Komsomol’skii pr. 29a, Perm, 614000 Russia Received March 21, 2006
DOI: 10.3103/S1066369X07060072
In this paper, we study the stability of a solution to the distributed delay equation t−τ x(s)ds = f (t), t ∈ R+ ; x(t) ˙ + ax(t) + b t−τ −h
x(ξ) = 0,
(1)
ξ < 0,
where the parameters a, b, τ , and h are assumed to be constant. The stability of linear functional differential equations is considered in many papers. Note that most of them deal with equations with a concentrated delay. For these equations the efficient stability tests are proposed (see [1–7] and references therein). The amount of stability tests for equations in form (1) is much less. In paper [8], an asymptotic stability criterion is obtained for a solution to Eq. (1) with a = 0 and real b (evidently, for the first time). In papers [9–12] the sufficient stability conditions are formulated for Eqs. (1) with variable coefficients a, b, and τ = 0. Inserting the constant coefficients into their theorems, one obtains the tests which are far from the exact ones. The objective of this paper is to formulate the asymptotic stability criteria for solutions to Eq. (1) in terms of the coefficients of the initial problem. Let N = {1, 2, . . . }, N0 = {0, 1, 2, . . . }, R = (−∞, +∞), R+ = [0, +∞); let C be the space of complex numbers. Definition 1. We understand a solution to Eq. (1) as a function x : R+ → C which is absolutely continuous on any finite segment and satisfies (1) almost everywhere. It is well known ([1], P. 84, theorem 1.1) that Eq. (1) is uniquely solvable and its solution admits the representation t C(t, s)f (s)ds, (2) x(t) = C(t, 0)x(0) + 0
where the function C(t, s) is called the Cauchy function for Eq. (1) ([1], P. 84). Consider the function x0 : R+ → C which is a solution to the problem t−τ x0 (s)ds = 0, t ∈ R+ ; x˙ 0 (t) + ax0 (t) + b t−τ −h
x0 (0) = 1,
x0 (ξ) = 0,
(3)
ξ < 0.
We call the mentioned function a fundamental solution ([1], P. 34). Note the following fact: since a, b, τ , and h are independent of t, the functions x0 (t) and C(t, s) are connected by the relation x0 (t − s) = C(t, s). *
E-mail:
[email protected].
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(4)
SEVERAL STABILITY TESTS FOR LINEAR
53
According to (2) and (4), a solution to Eq. (1) is completely defined by the fundamental solution x0 (t). So further we study the properties of the function x0 (t). Among them the most important for us is the property of the exponential stability. Definition 2 ([1], pp. 89–90). We say that Eq. (1) is exponentially stable, if certain positive N and γ with any positive t meet the bound |x0 (t)| ≤ N e−γt .
(5)
Note that due to equality (4) condition (5) takes place if and only if the Cauchy function admits the exponential bound, what, in turn, is equivalent to the stability of Eq. (1) with respect to the right-hand side from the space Lp (with p > 1) ([2], theorem 3.3.1). Applying the Laplace transform to problem (3), we obtain X(p) = (g(p))−1 , b g(p) = p + a + e−pτ (1 − e−ph ). p
(6)
According to the formula of the inverse Laplace transform, λ+iy −1 1 lim ept g(p) dp; x0 (t) = 2πi y→∞ λ−iy
(7)
here the integral is taken along any straight line λ = Re p so that ept X(p) is analytical in the domain Re p > λ. Such a domain exists in view of the bound for the fundamental solution to Eq. (1) ([2], P. 98, property 2) |x0 (t)| ≤ e(|a|+|b|h)t . Note that the function g(p) has a removable singularity at p = 0, but for p = 0 the function g(p) is analytical. Consequently, X(p) has singularities (poles) only at the points, where the function g(p) = 0. Let us establish for Eq. (1) efficient tests of the exponential stability, imposing certain additional conditions on the coefficients a and b. Case I. a = 0, b = βeiψ ∈ C. Putting a = 0 in equalities (1) and (6), we obtain ⎧ t−τ ⎪ ⎨x(t) x(s)ds = f (t), ˙ +b ⎪ ⎩
t−τ −h
t ∈ R+ ;
(8)
x(ξ) = 0,
ξ < 0, b g(p) = p + e−pτ (1 − e−ph ). p
(9) pt
e . In order to prove these Let us mention two simple properties of the integrals of the function g(p) properties, it suffices to estimate the integrals along the straight lines which are parallel, correspondingly, to the imaginary axis and the real one.
Lemma 1. Any real μ admits the bound
μ+i∞ pt
e
≤ N eμt .
dp
μ−i∞ g(p) Lemma 2. Any real λ and μ such that μ < λ < 0 satisfy the relation
λ+iy pt
e
dp
−−−−→ 0.
y→±∞ μ+iy g(p) RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 51 No. 6 2007
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The following lemma describes one property of function (9) which is most important for our study. Lemma 3. All zeros of function (9) lie to the left of the imaginary axis if and only if ψ ∈ (−π/2, π/2) and the following inequality is true: −1 π/2 − |ψ| 2 βh2 π/2 − |ψ| < 0< . (10) sin 2 2τ /h + 1 2τ /h + 1 Proof. Sufficiency. Let us use the principle of argument ([13], P. 82). Let p move along the contour KR of the complex plane which is composed of the semicircle CR : p = Reiy , −π/2 ≤ y ≤ π/2, and the segment IR : p = −iy, −R ≤ y ≤ R, R → ∞. Consider Arg g(−iy), where v(y) = −y + 2(β/y) sin(yh/2) sin(ψ + y(τ + h/2)), u(y) = 2(β/y) sin(yh/2) cos(ψ + y(τ + h/2)).
(11)
Assume that the hodograph curve does not go through the origin of coordinates. Then function (9) has no zeros at the right of the imaginary axis if and only if Δ Arg g(p) = 0. Evidently, with sufficiently large R, moving along the semicircle, we get Δ Arg g(p) = π. Note that Δ Arg g(0) = ψ; let us prove that (10) yields Δ Arg g(p) = −ψ − π/2, when the variable p moves along the imaginary semiaxis 0 ≤ y < ∞. Hereinafter, if the contrary is not stated, we define the increment of the argument of the function for y, varying from 0 to ∞. Let (yj ) be a monotonically increasing sequence of zeros of the function u: for any j∈N, u(yj ) = 0, yj+1 > yj > 0; let vj = v(yj ). Since lim v(y) = −∞ and u is a bounded function of y, the assertion y→+∞
that vj < 0 for all j ∈ N yields Δ Arg g(−iy) = −π/2. Consider the limit position of the hodograph curve g(−iy), i. e., a position such that the hodograph curve first intersects the imaginary axis at the origin of coordinates. According to (11), in this case, −1 2 π/2 − |ψ| 2 2 π/2 − |ψ| π/2 − |ψ| and β = 2 . sin y1 = h 2τ /h + 1 h 2τ /h + 1 2τ /h + 1 Let us prove that (10) implies that vj < 0 for any j ∈ N. The hodograph curve intersects the imaginary , n ∈ N0 , as well as yk∗∗ = 2πk/h, k ∈ N. Denote v(yn∗ ) = vn∗ , axis at points in the form yn∗ = π(1+2n)−2ψ 2τ +h v(yk∗∗ ) = vk∗∗ . Evidently, vk∗∗ < 0 for any k ∈ N, β, h, and τ . The condition vn∗ < 0 is equivalent to that (π(1 + 2n) − 2ψ)2 (π/2)(1 + 2n) − ψ n < , n ∈ N0 . (12) (−1) sin 2τ /h + 1 2β(2τ + h)2 > 0 (one can consider the opposite inequality analogously). Let, for example, sin (π/2)(1+2n)−ψ 2τ /h+1 Then, evidently, for odd n inequality (12) is true. If n = 0, then (12) immediately follows from the condition of Lemma 1. For n = 0, using the conditions of the lemma, we obtain
(π/2)(1+2n)−ψ 2 sin 4(π/2 − |ψ|) 2τ /h+1 (π/2)(1 + 2n) − ψ
< 2 = A. β sin 2 2τ /h + 1 2h (2τ /h + 1) sin π/2−|ψ| 2τ /h+1
Heeding the condition ψ ∈ (−π/2, π/2), we have
π/2−|ψ| 2τ /h+1
(π/2)(1+2n)−ψ > π/2−|ψ| 2τ /h+1 2τ /h+1 sin x/x2 is decreasing with
∈ (0, π/2] and
with any h > 0, τ > 0, n ∈ N. On the other hand, evidently, the function x ∈ (0, 3π/2) and sin ξ/ξ 2 > sin ζ/ζ 2 for ξ ∈ (0, π/2) and ζ ∈ [π/2, ∞). Therefore, A
0 and γ > 0 such that for all t and s, satisfying the inequality t ≥ s ≥ 0, the Cauchy function for Eq. (1) admits the bound |C(t, s)| ≤ N e−γ(t−s) ; c) the coefficient b = βeiψ meets the inequalities −π/2 < ψ < π/2 and (10). In Fig. 1, the exponential stability domain for Eq. (8) is constructed in terms of the parameters bh2 , τ . Evidently, it is symmetric with respect to the real axis, and its boundary is smooth, except for the points which belong to the domain for ψ = 0, but bh2 = 0.
Fig. 1.
One can easily generalize Theorem 1 for the case of a system of linear autonomous differential equations with a distributed delay t−τ x(s)ds = 0, t ∈ R+ ; x(t) ˙ +B (15) t−τ −h x(ξ) = 0, ξ < 0. Theorem 2 ([14]). A system of differential equations in form (15) is exponentially stable if and only if any eigenvalue λ = |λ|eiψ of the matrix B meets the inequalities −π/2 < ψ < π/2 and (10) with β = |λ|. Let us compare the criterion obtained in Theorem 1 with the known results. Rewrite Eq. (8), putting b = ε/h and f (t) ≡ 0, ε t−τ x(s)ds = 0, t ∈ R+ . (16) x(t) ˙ + h t−τ −h If ε is a real value, then Theorem 1 implies the well-known asymptotic stability test for a scalar differential equation with a distributed delay. Corollary 1 ([8]). Eq. (16) is exponentially stable if and only if πh 2 2 . 2(2τ + h) sin 0 0 and γ > 0 such that for all t and s, satisfying the inequality t ≥ s ≥ 0, the Cauchy function for Eq. (1) admits the bound |C(t, s)| ≤ N e−γ(t−s) ; c) the parameters a, b, h, τ of Eq. (1) are such that {ah, bh2 , τ /h} ∈ D1 . Let us compare this result with the known ones. In Eq. (1) we again put b = ε/h and f (t) ≡ 0: ε t−τ x(t) ˙ + ax(t) + x(s)ds = 0, t ∈ R+ . (17) h t−τ −h If ε is a real value, then proceeding in (17) to the limit for h → 0, from Theorem 3 we obtain the well-known asymptotic stability test for a scalar differential equation with a concentrated delay. Corollary 3 ([5], P. 57; [16]). The equation x(t) ˙ + ax(t) + εx(t − τ ) = 0 is exponentially stable if and only if the parameters a, ε, τ are such that {aτ, ετ } ∈ D, where the domain D contains the positive semiaxis aτ and is bounded by the lines a + ε = 0 and aτ = −θ cot θ, ετ = θ/ sin θ (0 ≤ θ < π). Case III. a = αeiϕ ∈ C, b ∈ R. The surface Γ2 =
⎧ 2θ ⎨αh = − 2τ /h+1 ⎩bh2 =
cos θ sin(ϕ+θ) , 2θ 2 cos ϕ , θ 2 sin(ϕ+θ) (2τ /h+2) sin 2τ /h+1
θ ∈ R,
divides the space of the parameters ah, bh2 , τ /h onto separate open sets. Let D2 stand for the set, containing the exponential stability domain for a solution to problem (8) with b ∈ R, as well as the exponential stability domain for a solution to an ordinary differential equation (b = 0) (it does not contain Γ2 ). See Fig. 3 for the domain D2 with fixed bh2 . Repeating the reasoning of Lemma 5 and proving the analogs of lemmas 1, 2, and 4, we obtain the final result. Theorem 4. The following statements are equivalent: a) Eq. (1) is exponentially stable; b) with certain positive N and γ for any t and s such that t ≥ s ≥ 0 the Cauchy function admits the bound |C(t, s)| ≤ N e−γ(t−s) ; c) the parameters a, b, h, τ of Eq. (1) are such that {ah, bh2 , τ /h} ∈ D2 . In paper [4] (P. 92, corollary 3.4) an asymptotic stability test is obtained for the equation x(t) ˙ + ax(t) + εx(t − τ ) = 0, t ∈ R+ , with complex coefficients a and ε. According to the said above, one can obtain this equation, proceeding in (17) to the limit for h → 0. Let us put b = ε/h in theorems 3 and 4 and proceed to the limit for h → 0. Then this criterion becomes equivalent to the stability tests obtained in the mentioned theorems. ACKNOWLEDGMENTS The work was supported by the Russian Foundation for Basic Research - Ural (project № 01-96069). RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 51 No. 6 2007
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Fig. 3.
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