PROCEEDINGS OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 124, Number 5, May 1996
ON THE PROBLEM OF LINEARIZATION
FOR STATE-DEPENDENT DELAY DIFFERENTIAL EQUATIONS
KENNETH L. COOKE AND WENZHANG HUANG
(Communicated by Hal L. Smith)
Abstract. The local stability of the equilibrium for a general class of statedependent delay equations of the form
ẋ(t) = f
Z
0
xt ,
dη(s)g(xt (−τ (xt ) + s))
−r0
has been studied under natural and minimal hypotheses. In particular, it
has been shown that generically the behavior of the state-dependent delay τ
(except the value of τ ) near an equilibrium has no effect on the stability, and
that the local linearization method can be applied by treating the delay τ as
a constant value at the equilibrium.
1. Introduction
Differential delay equations or functional differential equations have been used
in the modeling of scientific phenomena for many years. Often it has been assumed
that the delays are either fixed constants or are given as integrals (distributed
delays). However, in recent years, the more complicated situation in which the
delays depend on the unknown functions has been proposed in models [1, 2, 3, 4,
5, 6, 7, 9, 10, 11, 12, 13]. These equations are frequently called equations with
state-dependent delay. For example, it is a well-known fact that many types of
structured population models can be reduced to the study of functional differential
equations and in some cases this reduction yields a delay equation of threshold
type. For examples of these models and equations, see the references below.
Although state-dependent equations often are functional differential equations
of standard type, there are instances in which they are not, see [13], or in which
it would be useful to have results specific to this situation. The purpose of this
note is to show how to obtain a formal linearization around an equilibrium of a
quite general type of equation with state-dependent delay, and to give a proof of
this result under natural and minimal hypotheses. As will be shown, the result
is easy to understand and use. The result then provides the means to study the
Received by the editors December 3, 1993.
1991 Mathematics Subject Classification. Primary 34K20.
The first author’s research was supported in part by NSF grant DMS 9208818.
The second author’s research was supported in part by NSF grant DEB 925370 to Carlos
Castillo-Chavez and by the U.S. Army Research Office through the Mathematical Science Institute
of Cornell University.
c
1996
American Mathematical Society
1417
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1418
K. L. COOKE AND WENZHANG HUANG
asymptotic stability of an equilibrium of a state-dependent functional differential
equation, and the question of Hopf bifurcation from the equilibrium.
2. Formal linearization and stability
In this paper we consider the following type of state-dependent FDE:
(2.1)
x0 = φ ∈ C = C([−r, 0], Rn ),
ẋ(t) = F (xt ),
where xt ∈ C is defined by xt (s) = x(t + s), s ∈ [−r, 0], F takes the form
Z 0
dη(s)g(φ(−τ (φ) + s)) ,
F (φ) = f φ,
−r0
f : C × R → R , g : R → R are continuously differentiable and τ : C → [0, r1 ]
(r0 + r1 ≤ r) is continuous, and η is of bounded variation on [−r0 , 0]. The idea in
taking F (φ) of this form was suggested by the form in Alt [2] and the examples in
Section 4.
By the continuity of f , g and τ , one is able to see that F : C → Rn is continuous.
Hence following a standard approach we conclude the local existence of a solution to
Eq. (2.1). Nevertheless, if τ is not Lipschitzian, then F (φ) generally is not Lipschitz
continuous and hence the solution to (2.1) may not be unique. The purpose of this
paper is not to address the uniqueness of solution, but the local stability of an
equilibrium. As we can see later on in our analysis, nonuniqueness of solution does
not affect the local stability. The main reason is that a solution of (2.1) with φ
being an equilibrium is unique.
Without loss of generality we suppose
Z 0
dη(s)g(0) = 0,
F (0) = f 0,
n
n
n
n
−r0
that is, zero is a solution of (2.1). We note that F will not be differentiable at φ = 0
in general if τ is not differentiable at φ = 0. To study the local stability of the zero
solution we formally linearize (2.1) at zero by treating τ = τ (0) as a constant to
obtain a usual state-independent linear delay equation
Z 0
(2.2)
ẋ(t) = Dφ f (0, η̂g(0))xt + Dy f (0, η̂g(0))
dη(s)Dg(0)xt (−τ (0) + s),
−r0
R0
where η̂ = −r0 dη(s).
The purpose of this paper is to establish the following result which shows that
the stability of (2.2) does reflect the stability of the zero solution of the nonlinear
equation (2.1). That is, Eq. (2.2) behaves like a linearization of (2.1) at zero.
Theorem 2.1. Let the linear operator L : C → Rn be defined by
Z 0
(2.3)
Lφ = Dφ f (0, η̂g(0))φ + Dy f (0, η̂g(0))
dη(s)Dg(0)φ(−τ (0) + s).
−r0
(i) If sup{Reλ : det(λI − Le ) = 0} = −α < 0, then for each small ε > 0,
there exist a constant K(ε) and a neighborhood V (ε) ⊂ C of the origin such that
for φ ∈ V (ε), the solution xt (φ) of (2.1) is defined for t ≥ 0 and
λ·
kxt (φ)k ≤ K(ε)e−(α−ε)t kφk.
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STATE-DEPENDENT DELAY DIFFERENTIAL EQUATIONS
1419
(ii) If sup{Re λ : det(λI − Leλ· ) = 0} = α > 0, then the zero solution is
unstable.1
If the equilibrium solution is x∗ ∈ Rn instead of zero, so that
Z 0
F (x∗ ) = f x∗ ,
dη(s)g(x∗ ) = 0,
−r0
then the correct equation to replace (2.2) is
(2.20 )
ẋ(t) = Dφ f (x∗ , η̂g(x∗ ))xt + Dy f (x∗ , η̂g(x∗ ))
Z
0
−r0
dη(s)Dg(x∗ )xt (−τ (x∗ ) + s).
3. Proof of Theorem 2.1
Before proceeding to the proof of Theorem 2.1 we first introduce our notation
and give some preliminaries which are needed later on.
1. For notational simplicity, the Euclidean norm in Rn , supremum norm in C
and total variation of a function on [−r0 , 0] are all denoted by “k k”.
2. For h > 0, Bh := {φ ∈ C; kφk ≤ h}, Bhn := {y ∈ Rn ; kyk ≤ h}.
3. A solution of (2.1) is denoted by x(t, φ) or xt (φ) if it is considered in phase
space Rn or C, respectively.
Lemma 3.1. There exist δ > 0 and M > 0 such that
Z 0
f φ,
≤ M kφk,
(3.1)
dη(s)g(φ(−τ
(φ)
+
s))
−r0
φ ∈ Bδ .
Proof. Since f (φ, y) and g(z) are continuously differentiable in neighborhoods of
(0, η̂g(0)) in C × Rn and of 0 in Rn , respectively, there are δ1 > 0 and M1 > 0 such
that
kf (φ, y)k = kf (φ, y) − f (0, η̂g(0))k
≤ M1 (kφk + ky − η̂g(0)k),
φ ∈ Bδ1 , y − η̂g(0) ∈ Bδn1 ,
kg(z) − g(0)k ≤ M1 kzk,
Let
δ = min δ1 ,
δ1
M1 kηk
and M = M1 (1 + kηk).
Then for φ ∈ Bδ , we have
Z 0
Z
dη(s)g(φ(−τ (φ) + s)) − η̂g(0)
=
−r0
z ∈ Bδn1 .
0
−r0
dη(s)[g(φ(−τ (φ) + s)) − g(0)]
≤ kηkM1 kφk ≤ δ < δ1 .
1 We say that the zero solution is unstable if there is a δ > 0, such that for each ε > 0, there
is a φ ∈ C, kφk ≤ ε and tφ > 0 such that xt (φ) exists for 0 ≤ t ≤ tφ and kxtφ (φ)k ≥ δ.
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1420
Therefore
K. L. COOKE AND WENZHANG HUANG
Z
f φ,
dη(s)g(φ(−τ (φ) + s)) −r0
Z 0
≤ M1 kφk + dη(s)[g(φ(−τ (φ) + s)) − g(0)]
0
−r0
≤ (M1 + kηkM1 )kφk
= M kφk.
An immediate corollary of Lemma 3.1 is
Corollary 3.2. Let 0 < δ0 ≤ δ and 0 < δ ∗ < e−Mr δ0 be any fixed numbers. If
φ ∈ Bδ∗ , then x(t, φ) exists for t ∈ [−r, tφ ) with tφ > r. Furthermore, kxt (φ)k ≤
eMt kφk < δ0 , for t ∈ [0, r].
Proof. If the lemma is false for some x(t, φ), then by the continuity of x(t) = x(t, φ)
and the fact that kx0 k = kφk ≤ δ ∗ < δ0 , there must be a t∗ ∈ (0, r] such that
t ∈ [−r, t∗ ),
kx(t)k < δ0 ,
and kx(t∗ )k = δ0 .
Let x̂(t) = max−r≤s≤t {kx(s)k}, −r ≤ t ≤ t∗ ; then we have
0 ≤ x̂(t) < δ0 ,
(3.2)
t ∈ [−r, t∗ ),
x̂(t∗ ) = δ0 .
On the other hand, since x(t) satisfies the integral equation
Z t Z 0
x(t) = φ(0) +
f xs ,
dη(s)g(xs (−τ (xs ) + s)) ds,
−r0
0
by applying Lemma 3.1 we have
Z t Z
f xs ,
kx(t)k ≤ kφ(0)k +
Z
−r0
0
≤ kφk + M
0
t
kxs k ds,
dη(s)g(xs (−τ (xs ) + s)) ds
0 ≤ t ≤ t∗ .
0
It follows that
Z
x̂(t) ≤ kφk + M
(3.3)
t
x̂(s) ds.
0
Then the Gronwall inequality implies that
∗
∗
∗
kx(t∗ )k = x̂(t∗ ) ≤ kφkeMt ≤ δ ∗ eMt < δ0 e−M(r−t
)
≤ δ0 .
This contradicts (3.2).
Lemma 3.3. Let
G(ψ) = F (ψ) − Lψ.
Then there exists a constant N > 0 such that for each ν > 0 (ν ≤ 1), there is an
h(ν) > 0 such that for ψ ∈ Bh(ν) with ψ differentiable, we have
kG(ψ)k ≤ νN (kψk + kψ̇k).
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STATE-DEPENDENT DELAY DIFFERENTIAL EQUATIONS
1421
R0
Proof. For any fixed ψ ∈ C, let H(θ) = −r0 dη(s)g(θψ(−τ (ψ) + s)) and W (θ) =
f (θψ, H(θ)), θ ∈ [0, 1]. Then W (1) = F (ψ) and W (0) = 0. Therefore
(3.4)
F (ψ) =W (1)
Z 1
=
Ẇ (θ) dθ
0
Z
1
=
Dφ f (θψ, H(θ))ψ dθ
0
Z
+
Z
1
Dy f (θψ, H(θ))
0
0
dη(s)Dg(θψ(−τ (ψ) + s))ψ(−τ (ψ) + s) dθ.
−r0
By (1.3) and (3.4) it is not difficult to verify that
G(ψ) =
4
X
Gi (ψ),
i=1
with
Z
1
[Dφ f (θψ, H(θ)) − Dφ f (0, η̂g(0))]ψ dθ,
Z 1
G2 (ψ) =
[Dy f (θψ, H(θ)) − Dy f (0, η̂g(0))]
G1 (ψ) =
0
0
Z
·
Z
1
G3 (ψ) =
·
G4 (ψ) =
−r0
Z
1
dη(s)Dg(θψ(−τ (ψ) + s))ψ(−τ (0) + s) dθ,
Dy f (0, η̂g(0))
0
Z
0
0
−r0
dη(s)[Dg(θψ(−τ (ψ) + s)) − Dg(0)]ψ(−τ (0) + s) dθ,
Z
Dy f (θφ, H(θ))
0
0
dη(s)Dg(θψ(−τ (ψ) + s))
−r0
· [ψ(−τ (ψ) + s) − ψ(−τ (0) + s)] dθ.
By continuous differentiability of f and g and continuity of τ for each ν > 0 (ν ≤ 1),
we can choose an h(ν) > 0, such that for φ, ψ ∈ Bh(ν) ,
Z
Dφ f φ,
(3.5)
dη(s)g(φ(−τ (ψ) + s)) − Dφ f (0, η̂g(0)) ≤ ν,
−r0
Z 0
Dy f φ,
dη(s)g(φ(−τ (ψ) + s)) − Dy f (0, η̂g(0)) ≤ ν,
0
−r0
kDg(φ) − Dg(0)k ≤ ν,
|τ (ψ) − τ (0)| ≤ ν.
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1422
K. L. COOKE AND WENZHANG HUANG
If follows from (3.5) that for ψ ∈ Bh(ν) with ψ continuously differentiable,
kG1 (ψ)k ≤ νkψk,
kG2 (ψ)k ≤ νkηk(kDg(0)k + 1)kψk,
(3.6)
kG3 (ψ)k ≤ νkηk kDy f (0, η̂g(0))k kψk,
kG4 (ψ)k ≤ ν(kDy f (0, η̂g(0))k + 1)(kDg(0)k + 1)kηk kψ̇k.
Now the lemma follows immediately from (3.6) if we set
N = max{1 + kηk(kDg(0)k + kDy f (0, η̂g(0))k + 1),
kηk(kDf (0, η̂g(0))k + 1)(kDg(0)k + 1)}.
Now we are in a position to prove Theorem 2.1.
Proof of Theorem 2.1. Let T (t) be the semigroup generated by the solution of the
linear system (2.2). Then for each given ε ∈ (0, α], there exists a constant K1 (ε) ≥ 1
such that
ε
kT (t)φk ≤ K1 (ε)e−(α− 2 )t kφk,
for t ≥ 0, φ ∈ C.
We choose ν ∈ (0, 1] and δ0 ∈ (0, δ] such that
ν≤
ε
,
4N (1 + M )K1 (ε)
δ0 ≤ min
1
h(ν)
δ
,
,
K1 (ε) M K1 (ε) K1 (ε)
,
where δ, M , N and h(ν) are defined as in Lemma 3.1 and Lemma 3.3. We now
claim that, for φ ∈ Bδ∗ with δ ∗ < e−Mr δ0 ,
(3.7)
kxt (φ)k < K1 (ε)δ0 ,
t ∈ [0, tφ ).
We prove this by contradiction. By Corollary 3.2 we have tφ > r and kxt (φ)k < δ0 ,
t ∈ [0, r]. So if inequality (3.7) is not true, then there must be a t∗ > r such that
(3.8)
kxt (φ)k < K1 (ε)δ0 ,
t ∈ [0, t∗ ),
and
kxt∗ (φ)k = K1 (ε)δ0 .
On the other hand, by the formula of variation of constants (see [8], Section 6.2, p.
173) for t ∈ [r, t∗ ] we have
xt (φ) = xt−r (xr (φ))
(3.9)
Z
= T (t − r)xr (φ) +
t−r
T (t − r − s)X0 G(xs+r (φ)) ds.
0
It follows from Lemma 3.1 and (3.8) that for t ∈ [0, t∗ ],
kẋ(t, φ)k = kF (xt (φ))k ≤ M kxt (φ)k ≤ M K1 (ε)δ0 .
Therefore, xt (φ) is differentiable for t ∈ [r, t∗ ) and
(3.10)
kẋt (φ)k ≤ M K1 (ε)δ0 ,
t ∈ [r, t∗ ].
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STATE-DEPENDENT DELAY DIFFERENTIAL EQUATIONS
1423
(3.9) and (3.10) imply that for t ∈ [r, t∗ ],
kxt−r (xr (φ))k
Z
t−r
≤ kT (t − r)k kxr (φ)k +
kT (t − r − s)k kG(xs+r (φ))k ds
0
ε
(3.11)
≤ K1 (ε)e−(α− 2 )(t−r) kxr (φ)k
Z t−r
ε
+ K1 (ε)νN
e−(α− 2 )(t−r−s) (kxs+r (φ)k + kẋs+r (φ)k) ds
0
ε
≤ K1 (ε)e−(α− 2 )(t−r) kxr (φ)k
Z t−r
ε
+ K1 (ε)νN
e−(α− 2 )(t−r−s) (1 + M )kxs (xr (φ))k ds.
0
By applying the Gronwall inequality we therefore have
kxt∗ (φ)k = kxt∗ −r (xr (φ))k
ε
∗
≤ K1 (ε)e−[α− 2 −νN (1+M)K1 (ε)](t
≤ K1 (ε)e
−(α−ε)(t∗ −r)
−r)
kxr (φ)k
δ0
< K1 (ε)δ0 .
The last inequality contradicts (3.8). This proves our claim. As a consequence
of (3.7) we have tφ = ∞ and that (3.7) is valid for all t ≥ r. By applying again the
Gronwall inequality to (3.11) and (3.3) we have
kxt (φ)k = kxt−r (xr (φ))k
(3.12)
≤ K1 (ε)e−(α−ε)(t−r) kxr (φ)k
≤ K1 (ε)e−(α−ε)(t−r) eMr kφk
= K1 (ε)e(α−ε+M)r e−(α−ε)t kφk,
t ≥ r,
and
(3.13)
kxt (φ)k ≤ eMt kφk = e(M+α−ε)r e−(α−ε)t kφk,
t ∈ [0, r].
If we let K(ε) = K1 (ε)e(α−ε+M)r , then (3.12) and (3.13) imply
kxt (φ)k ≤ K(ε)e−(α−ε)t kφk,
t ≥ 0, φ ∈ Bδ∗ .
This proves part (i) of Theorem 2.1. For part (ii), we only give an outline of the
proof. The idea of the proof is the same as for the state-independent delay FDE.
In fact, if α > 0, then by using the same estimate as in the proof for part (i) and a
similar argument to the usual FDE (for instance, see Theorem 1.1, Chapter 10 in
[8]), we are able to show that equation (2.1) has a nonzero solution x(t) which is
defined for all t ≤ 0 and
lim x(t) = 0.
t→−∞
Thus the zero solution is unstable.
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1424
K. L. COOKE AND WENZHANG HUANG
4. Examples
Example 1. Consider the equation
Z 0
Z
(4.1)
ẋ(t) = h
dη1 (s)x(t + s),
−r0
0
−r0
dη2 (s)x(t − τ (xt ) + s) ,
where x(t) ∈ R , h : R × R → R is differentiable, τ : C → R+ is continuous and
η1 , η2 are of bounded variation on [−r, 0]. Suppose that there is an x∗ ∈ Rn such
that
Z 0
Z 0
dη1 (s)x∗ ,
dη2 (s)x∗ = 0.
h
n
n
n
n
−r0
−r0
Then x∗ is an equilibrium of (4.1). By Theorem 2.1 the formal linearization at x∗
is
ẋ(t) = Lxt ,
with
Z
Lφ =D1 h
0
−r0
Z
+ D2 h
dη1 (s)x∗ ,
0
Z
−r0
∗
−r0
0
dη2 (s)x∗
Z
dη1 (s)x ,
0
−r0
Z
0
−r0
∗
Z
dη2 (s)x
dη1 (s)φ(s)
0
−r0
dη2 (s)φ(−τ (x∗ ) + s).
Here Di h, i = 1, 2, is the derivative of h with respect to the ith variable, respectively.
A simple equation of the form (4.1) is
ẋ(t) = Ax(t) + Bx(t − τ (xt )),
in which A and B are n × n matrices. In this case h takes the form
h(x, y) = Ax + By,
and
Z
Z
0
−r0
dη1 (s)φ(s) =
0
−r0
dη2 (s)φ(s) = φ(0).
It is clear that x∗ = 0 is an equilibrium and
Z 0
Z
Lφ = D1 h(0, 0)
dη1 (s)φ(s) + D2 h(0, 0)
−r0
0
−r0
dη2 (s)φ(−τ (x∗ ) + s)
= Aφ(0) + Bφ(−τ (0)).
Hence the formal linearization at zero is
ẋ(t) = Ax(t) + Bx(t − τ (0)).
Example 2. Consider the functional equation
Z t
(4.2)
g(x(s)) ds,
x(t) =
t−ξ(x(t))
where g : R → R and ξ : R → R are continuously differentiable. (4.2) has been
studied in [3]. By differentiating (4.2) we can obtain a differential equation
+
(4.3)
ẋ(t) =
g(x(t)) − g(x(t − ξ(x(t))))
.
1 − ξ 0 (x(t))g(x(t − ξ(x(t))))
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STATE-DEPENDENT DELAY DIFFERENTIAL EQUATIONS
1425
Note that equation (4.3) is not equivalent to (4.2). It is clear that every solution of
(4.2) is a solution of (4.3) but the reverse is not true. In fact any constant function
is a solution of (4.3) but clearly it may not necessarily be a solution of (4.2).
Nevertheless we can use (4.3) to study the stability for equation (4.2). Suppose
that x is an equilibrium of (4.2), that is, x = g(x)ξ(x). We formally linearize (4.3)
around x. Now (4.3) is a special case of (4.1) with
h(x, y) =
and
Z
Z
0
−r0
g(x) − g(y)
,
1 − ξ 0 (x)g(y)
dη1 (s)φ(s) =
0
−r0
dη2 (s)φ(s) = φ(0).
Therefore we have the corresponding linear system at x:
ẋ(t) = hx (x, x)x(t) + hy (x, x)x(t − ξ(x))
(4.4)
g 0 (x)
=
[x(t) − x(t − ξ(x))].
1 − g(x)ξ 0 (x)
As we mentioned above, the stability of the zero solution of (4.4) is not equivalent
to the stability of x as an equilibrium of (4.2) ((4.4) always has a zero eigenvalue!).
However, we have2
The equilibrium x is asymptotically stable if zero is a simple eigenvalue, and all
other eigenvalues of (4.4) have negative real part.
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1426
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Department of Mathematics, Pomona College, Claremont, California 91711
E-mail address: [email protected]
Department of Mathematics, University of Alabama in Huntsville, Huntsville, Alabama 35899
E-mail address: [email protected]
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