Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 210, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
ASYMPTOTIC BEHAVIOR OF SOLUTIONS TO MIXED TYPE
DIFFERENTIAL EQUATIONS
SANDRA PINELAS
Abstract. This work concerns the asymptotic behavior of solutions to the
differential equation
ẋ(t) +
m
X
ai (t)x(ri (t)) +
i=1
n
X
bj (t)x(τj (t)) = 0,
j=1
where aj (t) and bj (t) are real-valued continuous functions and rj (t) and τj (t)
are non-negative functions such that
ri (t) ≤ t, t ≥ t0 ,
τj (t) ≥ t, t ≥ t0 ,
lim ri (t) = ∞, i = 1, . . . , m;
t→∞
lim τj (t) = ∞, j = 1, . . . , n.
t→∞
1. Introduction
In recent years, the theory of delay differential equations with advanced and
retarded arguments (mixed type) has provided a natural framework for mathematical modeling of many real world phenomena, namely optimal control problems [6],
nerve conduction theory [3], the slowing down of neutrons in nuclear reactors [9],
models for economic dynamics [6, 7] and the description of traveling waves in a
spatial lattice [4, 5]. See Bellman and Cooke [1] for more applications of differential
equations of mixed type. The concept of delay is related to the memory of systems,
where past events influence the current behavior. The concept of advance is related
to a potential future events which are known at the current time, and which could
be useful for decision making. It is well known that the solutions of these types of
equations cannot be obtained in closed form. In the absence of a closed form, a
viable alternative is studying the qualitative behavior of solutions. As a first step,
we need existence and uniqueness of solutions which can be a complicated issue for
mixed type equations.
In this article we study the asymptotic behavior of the advanced and retarded
differential equation
0
x (t) +
m
X
i=1
ai (t)x(ri (t)) +
n
X
bj (t)x(τj (t)) = 0,
t ≥ t0 ,
j=1
2000 Mathematics Subject Classification. 34K06, 34D04, 47H10.
Key words and phrases. Differential equations; fixed point; asymptotic behavior.
c
2014
Texas State University - San Marcos.
Submitted May 5, 2014. Published October 10, 2014.
1
(1.1)
2
S. PINELAS
EJDE-2014/210
where ai (t) and bj (t) are real continuous functions and rj (t) and τj (t) are nonnegative functions such that
ri (t) ≤ t, t ≥ t0 ,
τj (t) ≥ t, t ≥ t0 ,
lim ri (t) = ∞, i = 1, . . . , m;
t→∞
lim τj (t) = ∞, j = 1, . . . , n.
t→∞
About fifteen years ago, a new technique of fixed points was developed for studying stability of delay differential equations [2]. In the present article we apply this
technique to mixed type differential equation. It is possible to find in the literature
some conditions to ensure the stability of a solution of a delay differential equation,
but it is not easy to find conditions for stability for mixed type differential equations. We will establish necessary and sufficient conditions for all solutions of (1.1)
to converge to zero.
In the second section we establish the main results, and in the third section we
illustrate the previous results with an example.
2. Main results
Let r0 = inf{ri (s) : s ≥ t0 , i = 1, . . . , m}. Then r0 ≤ t0 , and the initial condition
for (1.1) is determined by a function φ, continuous on [r0 , t0 ],
x(t) = φ(t)
for r0 ≤ t ≤ t0 .
(2.1)
For short notation, we write x0 = x(t0 ) and y0 = y(t0 ).
By a solution to (1.1) we mean a continuous function x : [r0 , ∞) → R satisfying
(2.1), and differentiable on [t0 , ∞) and satisfies (1.1).
Theorem 2.1. Let ai (t) and bj (t) non-positive functions. Suppose that the inequality
y(t) ≥ −
m
X
−
ai (t)e
Rt
ri (t)
y(s) ds
i=1
−
n
X
bj (t)e
R τj (t)
t
y(s) ds
for t ≥ t0
(2.2)
j=1
has a nonnegative solution which is integrable on each interval [t0 , b]. Then (1.1)
has a positive solution.
Proof. Let y0 (t) be a nonnegative solution of (2.2). Define the iteration
(
yk (t)
r0 ≤ t ≤ t0 ,
R
R τj (t)
yk+1 (t) =
Pm
Pn
− rt (t) yk (s) ds
yk (s) ds
− i=1 ai (t)e i
− j=1 bj (t)e t
, t0 ≤ t
for k = 0, 1, . . . . Then, by (2.2), we have
y1 (t) = −
m
X
ai (t)e
i=1
−
Rt
ri (t)
y0 (s) ds
−
n
X
bj (t)e
R τj (t)
t
y0 (s) ds
≤ y0 (t) .
j=1
By induction we have 0 ≤ yk+1 (t) ≤ yk (t) ≤ · · · ≤ y0 (t). Hence, there exists a
pointwise limit y(t) = limk→∞ yk (t). By the Lesbesgue convergence theorem, we
have
m
n
R
R τj (t)
X
X
− t
y(s) ds
y(s) ds
y(t) = −
ai (t)e ri (t)
−
bj (t)e t
.
i=1
j=1
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ASYMPTOTIC BEHAVIOR OF SOLUTIONS
3
Then, the function
(
Rt
y(s) ds
y(t0 )e t0
x(t) =
y(t)
t ≥ t0 ,
r0 ≤ t ≤ t0
is a positive solution of (1.1).
Theorem 2.2. Let ai (t) and bj (t) be non-positive functions. If
Z ∞X
n
bj (s) ds = −∞
t0
j=1
and x(t) is a non-oscillatory solution of (1.1), then limt→∞ x(t) = ∞.
Proof. Since x is non-oscillatory, it must be eventually positive or eventually negative. We consider only the positive case, because in the negative case we can
consider −x which is also a solution. Suppose that x(t) > 0 for t ≥ t01 . Select
t1 ≥ t01 such that t01 ≤ inf{ri (s) : s ≥ t1 , i = 1, . . . , m}. Then x0 (t) ≥ 0 for t ≥ t1 ,
and
m
n
X
X
x0 (t) = −
ai (t)x(ri (t)) −
bj (t)x(τj (t))
i=1
≥−
n
X
j=1
bj (t)x(τj (t))
j=1
≥ −x(t1 )
n
X
bj (t),
j=1
which implies
x(t) ≥ −x(t1 )
Z tX
n
bj (s) ds.
t0 j=1
Thus, limt→∞ x(t) = ∞.
Theorem 2.3. Let ai (t) and bj (t) non-negative functions. If, either
Z ∞X
Z ∞X
n
n
ai (s) ds = ∞ or
bj (s) ds = ∞ ,
t0
t0
i=1
j=1
and x(t) is a non-oscillatory solution of (1.1), then limt→∞ x(t) = 0.
Proof. Suppose that x(t) > 0 for t ≥ t1 . Select t1 ≥ t01 such that t01 ≤ inf{ri (s) :
s ≥ t1 , i = 1, . . . , m}. Then x0 (t) ≤ 0 for t ≥ t1 . thus, x(t) is non-increasing and
positive. It must have a finite limit. If limt→∞ x(t) = d > 0, then x(t) > d for
t ≥ t1 , and
m
n
X
X
x0 (t) ≤ −d
ai (s) +
bj (t)
i=1
j=1
which implies limt→∞ x(t) = −∞. This contradicts to the assumption that x(t) is
positive, and therefore limt→∞ x(t) = 0.
Next we study the asymptotic behavior of (1.1), independently of the sign of the
coefficients. In the next lemma we establish an equivalence between the differential
equation (1.1) and an integral equation.
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S. PINELAS
EJDE-2014/210
Lemma 2.4. A function x(t) is a solution of (1.1) and (2.1) if and only if x(t) is
a solution of
Z t R
R
t
− tt (A(s)+B(s)) ds
−
e− u (A(s)+B(s)) ds
x(t) = x0 e 0
t0
×
m
X
Z
u
Ex (s) ds −
ai (u)
ri (u)
i=1
n
X
Z
τj (u)
bj (u)
(2.3)
Ex (s) ds du
u
j=1
for t ≥ t0 , and (2.1) is satisfied. Here we use the notation
m
X
A(t) =
ai (t),
B(t) =
i=1
Ex (t) =
m
X
m
X
bj (t),
j=1
ai (t)x(ri (t)) +
i=1
n
X
bj (t)x(τj (t)).
j=1
Proof. Note that by (1.1), x0 (t) = −Ex (t) and that
Z t
Z
0
x(ri (t)) = x(t) −
x (u) du = x(t) +
ri (t)
τj (t)
Z
x(τj (t)) = x(t) +
t
Ex (u) du,
ri (t)
x0 (u) du = x(t) −
t
Z
τj (t)
Ex (u) du .
t
Then (1.1) can be re-written as
Z t
Z
m
n
X
X
x0 (t) +
ai (t) x(t) −
ẋ(s) ds +
bj (t) x(t) −
ri (t)
i=1
t
ẋ(s) ds = 0
τj (t)
j=1
which is equivalent to
x0 (t) + (A(t) + B(t))x(t) = −
m
X
Z
t
ai (t)
Ex (s) ds +
ri (t)
i=1
n
X
Z
bj (t)
τj (t)
Ex (s) ds.
t
j=1
(2.4)
Rt
Multiplying both sides by the integrating factor exp( t0 (A(s)+B(s)) ds), we obtain
differential equation equivalent to the one above:
d Rtt (A(s)+B(s)) ds
e 0
x(t)
dt
Z t
Z τj (t)
n
m
X
Rt
X
(A(s)+B(s)) ds
t0
= −e
ai (t)
Ex (s) ds −
bj (t)
Ex (s) ds .
ri (t)
i=1
j=1
t
Integrating from t0 to t, we obtain (2.3).
Now starting from (2.3) differentiate with respect to t, and retrace the steps
above to obtain (1.1). The proof is complete.
Theorem 2.5. Assume that there exists a constant c such that
Z t R
Z u X
m
m
n
X
X
t
e− u (A(s)+B(s)) ds
|ai (u)|
|ak (s)| +
|b` (s)| ds
t0
+
n
X
j=1
Z
|bj (u)|
i=1
m
τj (u) X
|ak (s)| +
u
k=1
ri (u)
n
X
`=1
k=1
`=1
|b` (s)| ds du ≤ c < 1 .
(2.5)
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ASYMPTOTIC BEHAVIOR OF SOLUTIONS
5
Then for each initial condition (2.1), there exists a unique solution to (1.1). Furthermore, if
Z t
(A(s) + B(s)) ds = ∞,
(2.6)
lim
t→∞
t0
then, every solution of (1.1) converges to zero.
Proof. Let C([t0 , ∞)) be the set of real-valued functions, continuous on [t0 , ∞).
Then this is a Banach space with the norm
kxk = sup |x(t)|.
t≥t0
Based on (2.3), for a function x ∈ C([r0 , ∞)), we define the operator


x(t) if r0 ≤ t ≤ t0 ,



R
Rt

Rt
 − tt0 (A(s)+B(s)) ds
x(t0 ) − t0 e− u (A(s)+B(s)) ds
e
(T x)(t) =
Ru
R τ (u)
Pm
Pn


×
ai (u) ri (u) Ex (s) ds − j=1 bj (u) u j Ex (s) ds du

i=1



if t0 ≤ t.
It is clear that T maps C([t0 , ∞)) into C([t0 , ∞)) and preserves the values of x(t)
for t ∈ [r0 , t0 ]. We will proof that T is a contraction.
Let x, y be two continuous function on [t0 , ∞), and satisfying the same initial
conditions (2.1). Then for t ≥ t0 , we have
|(T x)(t) − (T y)(t)|
≤e
−
Rt
(A(s)+B(s)) ds
t0
Z
t
e−
|(x(t0 ) − y(t0 ))| +
Rt
u
(A(s)+B(s)) ds
t0
×
+
m
X
Z
u
|ai (u)|
|Ex (s) − Ey (s)| ds du
ri (u)
i=1
n
X
Z
τj (u)
|bj (u)|
|Ex (s) − Ey (s)| ds du .
u
j=1
Since x(t) = y(t) for r0 ≤ t ≤ t0 , and
|Ex (s) − Ey (s)| ≤ |
m
X
ak (t)x(rk (t)) −
k=1
n
X
+|
`=1
≤
m
X
m
X
ak (t)y(rk (t))|
k=1
n
X
b` (t)x(τ` (t)) −
b` (t)y(τ` (t))|
`=1
|ak (t)||x(rk (t)) − y(rk (t))| +
k=1
n
X
|b` (t)||x(τ` (t)) − y(τ` (t))|,
`=1
by (2.5), we obtain
|(T x)(t) − (T y)(t)| ≤ ckx(t) − y(t)k .
Consequently, the operator T has a unique fixed point in C([t0 , ∞)), and this fixed
point satisfies (2.1).
Let
L = x ∈ C([t0 , ∞)) : lim x(t) = 0 ,
t→∞
6
S. PINELAS
EJDE-2014/210
which is a closed subspace of C([t0 , ∞)). Now, we claim that T (L) ⊂ L and that
T preserves the initial conditions (2.1). Indeed, for x ∈ L, we have
Z t R
R
t
− t (A(s)+B(s)) ds
+
e− u (A(s)+B(s)) ds
|(T x)(t)| ≤ |x0 |e t0
t0
m
X
×
u
Z
|ai (u)|
|Ex (s)| ds +
ri (u)
i=1
n
X
Z
τj (u)
|bj (u)|
|Ex (s)| ds du .
u
j=1
(2.7)
Note that by (2.6),
lim |x0 |e
Rt
−
t0
(A(s)+B(s)) ds
t→∞
= 0.
On the other hand, since x ∈ L, for each > 0 there exists t01 ≥ t0 such that
|x(t)| < /2 for all t ≥ t01 . Select t1 ≥ t01 such that t01 ≤ inf{ri (s) : s ≥ t1 , i =
1, . . . , m}. Then for t ≥ t1 ,
Z
t
e
Rt
−
u
(A(s)+B(s)) ds
t0
m
X
Z
|Ex (s)| ds du
ri (u)
i=1
Z
t1
≤
e−
Rt
u
(A(s)+B(s)) ds
t0
+
m
X
Z
t
Z
e−
Rt
u
(A(s)+B(s)) ds
t1
u
|ai (u)|
|Ex (s)| ds du
ri (u)
i=1
2
u
|ai (u)|
m
X
Z
m
X
u
|ai (u)|
rj (u)
i=1
|ak (s)| +
k=1
n
X
|b` (s)| ds du
`=1
We observe that the first term on the right-hand side approaches zero as t → ∞.
Then there exists t2 ≥ t1 such that
Z t1 R
Z u
m
X
t
e− u (A(s)+B(s)) ds
|ai (u)|
|Ex (s)| ds du < .
2
t0
ri (u)
i=1
Then by (2.5)
Z t R
Z
m
X
t
|ai (u)|
e− u (A(s)+B(s)) ds
t0
≤
i=1
+
2 2
Z
t
e−
Rt
u
(A(s)+B(s)) ds
t1
u
|Ex (s)| ds du
ri (u)
m
X
Z
u
|ai (u)|
ri (u)
i=1
m
X
k=1
Since is arbitrarily small,
Z t R
Z
m
X
t
lim
e− u (A(s)+B(s)) ds
|ai (u)|
t→∞
t0
i=1
t0
j=1
n
X
|b` (s)| ds du < .
`=1
u
|Ex (s)| ds du = 0.
(2.8)
ri (u)
By a similar process, we prove that
Z t R
Z
n
X
t
lim
e− u (A(s)+B(s)) ds
|bj (u)|
t→∞
|ak (s)| +
τj (u)
|Ex (s)| ds = 0 .
(2.9)
u
Therefore (T x)(t) → 0; i.e., the fixed point x = T x, satisfies limt→∞ x(t) = 0.
The above theorem provides sufficient conditions for the convergence of solutions
to zero. The next theorem provides necessary conditions.
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ASYMPTOTIC BEHAVIOR OF SOLUTIONS
7
Theorem 2.6. Assume that (2.6) holds, and
Z t
(A(s) + B(s)) ds > −∞.
lim inf
t→∞
(2.10)
t0
If all solutions of (1.1) converge to zero, then (2.7) holds.
Proof. For the shake of contradiction suppose that (2.7) does not hold; i. e.,
Z tn
(A(s) + B(s)) ds = α < ∞.
lim inf
t→∞
t0
Then by (2.10), α > −∞. Then there exists a sequence {tn } approaching ∞, such
that
Z tn
(A(s) + B(s)) ds = α .
lim
n→∞
t0
Let x be a solution such that x(t0 ) = x0 6= 0. then
lim x0 e
−
R tn
t0
(A(s)+B(s)) ds
= x0 eα 6= 0 .
n→∞
(2.11)
By Lemma 2.4, x(tn ) satisfies (2.3) with tn instead of t. By (2.8) and (2.9),
Z tn R
tn
e− u (A(s)+B(s)) ds
lim
n→∞
×
t0
m
X
i=1
Z
u
Ex (s) ds −
ai (u)
ri (u)
n
X
j=1
Z
τj (u)
bj (u)
(2.12)
Ex (s) ds du = 0 .
u
Since all solutions approach zero, by (2.3), (2.11) and (2.12), it follows that
0 = lim x(tn ) = x0 eα + 0 6= 0 .
n→∞
This contradiction competes the proof.
3. An example
In this section we provide an example to illustrate our results.
Example 3.1. Consider the equation
ẋ(t) + sin(t)e−(a−1)t/a x(t/a) + 1 − sin(t) e(b−1)t x(bt) = 0,
t>0
(3.1)
where a > b > 1 and b − 1 < (a − 1)/a. Note that r0 = t0 = 0 in this example,
so the initial condition (2.1) reduces to x(t0 ) = x0 . We want to check that all the
assumptions of Theorem 2.5 are satisfied.
Z t
(A(s) + B(s)) ds
0
Z t
=
(sin(t)e−(a−1)t/a + 1 − sin(t) e(b−1)t ) ds
0
e−αt
1
β
1
= 2
α sin(t) − cos(t) + eβt
− 2
sin(t) + 2
cos(t) + k,
α +1
β
β +1
β +1
where α = (a − 1)/a, β = b − 1 and
k=
1
1
1
+ + 2
.
α2 + 1 β
β +1
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S. PINELAS
EJDE-2014/210
Since
e−αt
(α sin(t) − cos(t)) → 0, and
α2 + 1
β
1
1
sin(t) + 2
cos(t)) → ∞,
eβt ( − 2
β
β +1
β +1
as t → ∞, it follows that (2.6) holds. Also note that
Z t R
Z u
n
t
−αs
βs
e− u sin(s)e +(1−sin(s))e ds | sin(u)|e−αu
0
| sin(s)|e−αs
u/a
Z
bu
o
+ |1 − sin(s)|e
(| sin(s)|e−αs + |1 − sin(s)|eβs ) ds du
ds + |1 − sin(u)|eβu
u
Z u
Z t R
n
t
−αs
βs
(e−αs
<
e− u (sin(s)e +(1−sin(s))e ) ds e−αu
βs
u/a
0
+ 2eβs ) ds + 2eβu
bu
Z
o
(e−αs + 2eβs ) ds du
u
t
n
e−αu/a − e−αu
eβu − eβu/a e−αu
+2
α
β
0
−αu
−αbu
βbu
βu o
e
−
e
e
−
e
+ 2eβu
+2
du
α
β
n
Rt
−αs
βs
e−αt/a − e−αt
eβt − eβt/a < te− 0 (sin(s)e +(1−sin(s))e ) ds e−αt
+2
α
β
−αt
−αbt
βbt
βt o
e
−e
e −e
+ 2eβt
+2
→ 0 as t → ∞ .
α
β
Z
e−
=
Rt
u
(sin se
−αs
+(1−sin s)eβs ) ds
Note that
te−
Rt
0
(sin(s)e−αs +(1−sin(s))eβs ) ds
→0
as t → ∞,
and
e−αt
e−αt/a − e−αt
α
+2
e−αt − e−αbt
eβt − eβt/a eβbt − eβt + 2eβt
+2
→0
β
α
β
as t → ∞, when a > b > 1 and b − 1 < (a − 1)/a. So, there exists an , 0 < < c/2,
such that
Rt
−(a−1)t/a
+(1−sin(t))e(b−1)t ) ds
< ,
e− 0 (sin(t)e
and
Z t
e−
Rt
u
(sin(s)e−αs +(1−sin(s))eβs ) ds
Z
n
| sin(u)|e−αu
0
+ |1 − sin(s)|e
u
| sin(s)|e−αs
u/a
βs
ds + |1 − sin(u)|e
βu
Z
bu
o
| sin(s)|e−αs + |1 − sin(s)|eβs ds du
u
<
Therefore, the conditions of Theorem 2.5 are satisfied, consequently all solutions of
(3.1) converge to zero. In fact the function is x(t) = x0 e−t which converges to zero,
for each initial condition x0 .
Note that condition (2.9) holds, so the conditions for Theorem 2.6 are also satisfied.
EJDE-2014/210
ASYMPTOTIC BEHAVIOR OF SOLUTIONS
9
Acknowledgments. The author wants to thank Professor Julio G. Dix for his
support and suggestions.
References
[1] R. Bellman, K. L. Cooke; Differential-Difference Equations, Academic Press, New York, 1963.
[2] T. A. Burton, T. Furumochi; Fixed points and problems in stability theory for ordinary and
functional differential equations, Dynam. Systems Appl. 10 (2001), no. 1, 89-116.
[3] H. Chi, J. Bell, B. Hassard; Numerical solution of a nonlinear advance-delay-differential equation from nerve conduction theory, J. Math. Biol. 24 1986 , 583-601.
[4] J. Mallet-Paret; The Fredholm alternative for functional differential equations of mixed type,
J. Dynamics Differential Equations 11 (1999), 1-47.
[5] J. Mallet-Paret; The global structure of traveling waves in spatially discrete dynamical systems,
J. Dynamics Differential Equations 11 (1999), 49-127.
[6] L. S. Pontryagin, R. V. Gamkreledze, E. F. Mischenko; The Mathematical Theory of Optimal
Processes, Interscience, New York, 1962.
[7] A. Rustichini; Functional differential equations of mixed type: The linear autonomous case,
J. Dynamics Differential Equations 1 (1989), 121-143.
[8] A. Rustichini; Hopf bifurcation for functional differential equations of mixed type, J. Dynamics
Differential Equations 1 (1989), 145-117.
[9] M. Slater, H. S. Wilf; A class of linear differential equations, Pacific J. Math. 10 (1960),
1419-1427.
Sandra Pinelas
Academia Militar, Departamento de Ciências Exatas e Naturais, Conde Castro Guimarães,
Amadora, Portugal
E-mail address: sandra.pinelas@gmail.com