Uniform Boundedness and Uniform Ultimate Boundedness for Functional Differential Equations

By G. Makay

(Southern Illinois University, USA) (Bolyai Institute, Szeged, Hungary)

1. Introduction.

Consider the functional differential equation

x′ = f(t, x_{t})  (1)

with finite delay, i.e. let C be the Banach space of continuous functions ϕ : [−h, 0] → R^{r} with the supremum norm ∥·∥, x_{t}(s) = x(t+s) for s ∈ [−h, 0]. Assume that f is continuous and locally Lipschitz, so there is a unique solution x(t, t_{0}, ϕ) with x_{t_{0}} = ϕ of the initial value problem for (1) which depends continuously on the initial data.

In this paper we study some conditions in connection with uniform boundedness (UB) and uniform ultimate boundedness (UUB). We give the definitions of some properties (called UB and UUB at a time t_{0}) that are weaker than UB and UUB. Then we prove theorems for finite and infinite delay equations using Liapunov functionals that ensure these properties. We also give examples to illustrate the connections and differences between the definitions. The importance of this work can be seen by noting that if solutions are UB and UUB at t_{0} and the right-hand side of the equation is T -periodic in t, then there is a T -periodic solution (see e.g. [1] for the finite and [2] for the infinite delay case). Also, there are several spaces that can be considered in the infinite delay case, He, Huang and Wang [3] show some of them and prove theorems using Liapunov functionals to ensure UB and UUB. Hering [4], [5] worked with spaces and methods to prove UB and UUB similar to those we are concerned with here.

2. Main results.

First we introduce our definitions of uniform boundedness and uniform ultimate boundedness. They can be found with t_{0} = 0 in [1] but we need them for arbitrarily fixed t_{0}.

Definition. We say that the solutions of (1) are uniform bounded at t_{0} (UB(t_{0})) if for all B_{1} > 0 there is a B_{2} > 0 such that t ≥ t_{0} and ϕ ∈ C with ∥ϕ∥ < B_{1} imply |x(t, t_{0}, ϕ)| < B_{2}.

Definition. We say that the solutions of (1) are uniform ultimate bounded at t_{0} (UUB(t_{0})) if there is a B > 0 such that for all B_{3} > 0 there is a K > 0 such that t ≥ t_{0} + K, ϕ ∈ C with ∥ϕ∥ < B_{3} imply |x(t, t_{0}, ϕ)| < B.

Note, that the only difference between these definitions and those in [1] is the initial time when we start the solutions. By a simple change of time one could prove that the solutions of x′ = f(t, x_{t}) are UB(t_{0}) (UUB(t_{0}), respectively) if and only if the solutions of x′ = f(t − t_{0}, x_{t}) are UB(0) (UUB(0)). It is also clear, that these conditions are generalizations of the properties known as uniform boundedness and uniform ultimate boundedness, since in those we have the constants B_{2}, B and K independent of t_{0} and the estimations for all t_{0}. With h = 0 these definitions work for ordinary differential equations as well.

Theorem 1: Suppose that f(t + T, ϕ) = f(t, ϕ) and for all B > 0 there is an L > 0 such that (t, ϕ) ∈ R×C with ∥ϕ∥ ≤ B implies |f(t, ϕ)| ≤ L. Assume also, that the solutions of (1) are UB and UUB at t_{0}. Then there is a T -periodic solution of (1).

Proof: The proof is the same as that of Theorem 4.2.2 in [1], the only difference is that instead of starting the solutions at 0 we need to start them at t_{0}. Actually, that theorem is the same as this one with t_{0} = 0 and does not use the stronger UB or UUB.

The condition on f we have in this theorem is called the boundedness condition. Note that, if f is locally Lipschitz, then we have this condition. We will clarify what kind of local Lipschitz condition we are talking about here, and how we can prove the boundedness condition needed for this theorem.

To see the point in these definitions we now give an example, where the solutions are not UB or UUB but they are UB and UUB at 0. To this end we need the function g : C × [−h, 0] × [−h, 0] → R defined by

g(ϕ, α, β) = sup_{α≤s_{0}≤...≤s_{n}≤β} ∑_{k=1}^{n} max{0, ϕ(s_{k}) − ϕ(s_{k−1}) − 1}.

For this definition to make sense we always assume that α ≤ β. Kato [6] defined a function g very similar to our function g and proved that it is continuous and locally Lipschitz in ϕ. It is also clear that g is continuous in α and β too. To clarify matters we note that g is a function, which gives the “large increasing variation” of ϕ on the interval [α, β]. We also define a function, which is (more or less) a sign function:

h(s) := \{ \begin{matrix}1, & if s ≥ 1 \\ s, & if s ∈ [−1, 1] \\ −1, & if s ≤ −1 \end{matrix}

Now we are ready to define f. Let t ∈ [2n, 2n + 2] for some integer n, then

f(t, ϕ) := \{ \begin{matrix}min\{g(ϕ, 2n − t, 0) − 1, 4|ϕ(0)|^{2}\}h(ϕ(0)), & if t ∈ [2n, 2n + 1] \\ min\{g(ϕ, −1, 2n + 1 − t) − 1, 4|ϕ(0)|^{2}\}h(ϕ(0)), & if t ∈ [2n + 1, 2n + 2] \end{matrix}

where h = 1 and T = 2. It is clear, that f is continuous and locally Lipschitz, since g and h are. But, unfortunately, we have two slightly different definitions of the local Lipschitz condition:

Definition. We say that a function f : R × C → R^{r} is weakly locally Lipschitz in its second variable, if for all (t_{0}, ϕ) ∈ R × C there is a δ > 0 and there is an L > 0 such that |t − t_{0}| < δ and ∥ψ − ϕ∥ < δ imply |f(t, ψ) − f(t, ϕ)| ≤ L∥ψ − ϕ∥.

Definition. We say that a function f : R × C → R^{r} is strongly locally Lipschitz in its second variable, if for all (t_{0}, ϕ) ∈ R × C and for all δ > 0 there is an L > 0 such that |t − t_{0}| < δ and ∥ψ − ϕ∥ < δ imply |f(t, ψ) − f(t, ϕ)| ≤ L∥ψ − ϕ∥.

The only difference in the two definitions is the way in which δ is defined. In his paper [6] Kato used the weak local Lipschitz condition when he said that g is locally Lipschitz, but unfortunately we need the strong one for f to prove the boundedness condition in Theorem 1. Note, that a weak local Lipschitz condition for f is enough to prove that

Uniform Boundedness and Uniform Ultimate Boundedness 4

solutions are unique and depend continuously on the initial data. We are unable to prove the strong local Lipschitz condition for this particular f, but we have the boundedness condition needed. That is why we put 4|ϕ(0)|^{2} into the argument of the minimum in the definition of f. Since 4|ϕ(0)|^{2} satisfies the boundedness condition, so does f. But 4|ϕ(0)|^{2} is a good enough right-hand side for a differential equation to ensure, that the solution blows up fast, so in our example we can be sure that the big influence of g will be in effect as we describe it in the following.

Let us consider how solutions of our example behave. First note that if x(t) = x(t, t_{0}, ϕ) ever becomes 0, then it remains 0 for ever after. Thus, solutions do not change sign. To be definite, consider a solution x(t) = x(t, t_{0}, ϕ) with ϕ(0) > 0, so x(t) ≥ 0 for all t ≥ t_{0}. Since g ≥ 0 and h ≤ 1, x′(t) ≥ −1. Define n to be the integer with t_{0} ∈ (2n−2, 2n]. From the definition of f we see that x′(2n) = −h(x(2n)) ≤ 0 and hence x(t) is monotone decreasing for t ∈ [2n, ∞). Moreover, if x(t) > 1 then x′(t) = −1 and x is a linear function, if x(t) ≤ 1 then x′(t) = −x(t) and x is an exponentially decreasing function. A similar proof works if ϕ(0) < 0. Therefore, if we start our solution at t_{0} = 2n then the above consideration proves, that the solutions are UB(t_{0}) (for every B_{1} > 0 let B_{2} := B_{1}) and UUB(t_{0}) (let B := 1, then for B_{3} > 0 we may choose K := B_{3} − 1).

But the solutions are not UB or UUB at any other point (we will prove it for only one), and hence they are not UB or UUB. To see this, we take a function ϕ which is oscillatory with amplitude greater than 1. If we start the solution at t_{0} = 2n + 1, our function g in the definition of f is applied to a segment containing [−1/2, 0] of ϕ while t is in [2n + 1, 2n + (3/2)] and it gives larger and larger numbers as ϕ has higher and higher frequency. (Note, that although g satisfies a weak local Lipschitz condition, ϕ may be bounded by 1, but g is not bounded for these ϕ’s, so g does not satisfy a strong local Lipschitz condition.) The minimum makes everything a little bit complicated, but one can argue as follows. From the above we know that g(x_{t}, −1, 2n + 1 − t) ≥ g(ϕ, −1/2, 0) for t ∈ [2n + 1, 2n + (3/2)], where x(t) = x(t, t_{0}, ϕ) is the solution. We can choose ϕ so that g(ϕ, −1/2, 0) > 1 and ϕ(0) > 1. From the definition of f it is clear that x(t) is monotone increasing on the interval [2n + 1, 2n + (3/2)]. Thus, in the following, we do not have to worry about the function h, since it is 1 along this solution on the intervals

5 G. Makay

considered here. We claim that the solution satisfies x′(t) = g(x_{t}, −1, 2n + 1 − t) − 1 for t ∈ [2n + (5/4), 2n + (3/2)]. If this is not true, then since g(x_{t}, −1, 2n + 1 − t) is monotone decreasing, but 4|x(t)|^{2} is monotone increasing we must have x′(t) = 4|x(t)|^{2} on the interval [2n+1, 2n+(5/4)]. But the solution of this equation (with t_{0} = 2n+1 and ϕ(0) > 1) blows up before 2n+(5/4), which is not possible, since x′ ≤ g(x_{t}, −1, 2n+1−t)−1 ≤ g(ϕ, −1, 0)−1 (a fixed constant for every ϕ) for t ∈ [2n + 1, 2n + (3/2)]. This contradiction shows, that x′(t) = g(x_{t}, −1, 2n + 1 − t) − 1 ≥ g(ϕ, −1/2, 0) − 1 for t ∈ [2n + (5/4), 2n + (3/2)]. Hence, the solution can be as large as one wants at t = 2n, because one can have any large number of the form g(ϕ, −1/2, 0) with bounded ϕ’s. Therefore, the solutions are not UB(t_{0}). Also, since the solution is very large at t = 2n and cannot decrease faster than a linear function with slope 1, we cannot find a suitable B in the definition of UUB(t_{0}).

The above example shows, that there are differential equations with solutions that are not UB or UUB but UB(t_{0}) and UUB(t_{0}) for some t_{0}. Since we also have |x(t, t_{0}, ϕ)| → 0 for all (t_{0}, ϕ) ∈ R × C, the example shows that dissipativeness (see e.g. [7]) does not imply UB or UUB. With some slight changes one can even show that dissipativeness does not imply UB or UUB at any t_{0}. Another interesting connection between these properties is shown in the following

Theorem 2: If the solutions of (1) are uniformly bounded, if they are uniformly ultimately bounded at \overline{t}_{0}, and if the right-hand side of (1) is T -periodic in t, then the solutions are uniformly ultimately bounded.

Proof: Let B > 0 be that of uniform ultimate boundedness at \overline{t}_{0}. Since (1) is T - periodic, using a translation argument it is easy to prove that the solutions are uniformly ultimately bounded at any \overline{t}_{n} = \overline{t}_{0} + nT with the same bound B. Moreover, K is independent of n in the uniform ultimate boundedness. Let B_{3} > 0 be given, define B_{4} to be the B_{2} of uniform boundedness for B_{1} = B_{3}. For this B_{4} we find a K > 0 from the uniform ultimate boundedness at \overline{t}_{n} (for any n). Now, let (t_{0}, ϕ) ∈ R × C be given with ∥ϕ∥ < B_{3}. Choose the smallest \overline{t}_{n} with t_{0} ≤ \overline{t}_{n}. From the uniform boundedness we have ∥x_{\overline{t}_{n}}(·, t_{0}, ϕ)∥ < B_{4} and hence from the uniform ultimate boundedness |x(\overline{t}_{n} +t, t_{0}, ϕ)| < B for t ≥ K. This proves that if (t_{0}, ϕ) ∈ R × C with ∥ϕ∥ < B_{3} then |x(t_{0} + t, t_{0}, ϕ)| < B for t ≥ K + T, which was to be shown.

Uniform Boundedness and Uniform Ultimate Boundedness 6

The next natural question is, how can we prove that solutions are UB and UUB at t_{0}. We modify Theorem 4.2.11 in [1] to fit our needs.

Theorem 3: Suppose there are a continuous functional V : R × C → R, a sequence t_{n} ∈ R with h ≤ t_{n} − t_{n−1} ≤ k for some k > 0 and constants a, b, M, U > 0 such that (i) W_{1}(|ϕ(0)|) ≤ V (t, ϕ) and V (t_{n}, ϕ) ≤ W_{2}(∥ϕ∥) (ii) V′(t, x_{t}) ≤ M (iii) V′(t, x_{t}) ≤ −a|x′(t)| − b for |x(t)| ≥ U, where W_{1}(r) → ∞ as r → ∞. Then the solutions of (1) are uniformly bounded and uniformly ultimately bounded at t_{n} for any n. Also, B_{2} of the uniform boundedness and B, K of uniform ultimate boundedness does not depend on n.

Proof: First we prove uniform boundedness. Clearly, the proof is the same for any n, and we will see that B_{2} does not depend on n, so we prove everything for t_{0}. Let B_{1} > 0 be given; we need to find a B_{2} > 0 such that ϕ ∈ C with ∥ϕ∥ < B_{1} implies |x(t, t_{0}, ϕ)| < B_{2} for t ≥ t_{0}. Define L ≥ B_{1} so that −aL + Mk < −α for some α ∈ (0, bh). Define B_{2} by

B_{2} := W_{1}^{−1} (W_{2}(U + L) + Mk (2 + \frac{W_{2}(U + L)}{α}))

and suppose for contradiction that |x(t)| = |x(t, t_{0}, ϕ)| ≥ B_{2} for some t ≥ t_{0}. Using (i) we find that V (t) := V (t, x_{t}) ≥ W_{1}(|x(t)|) ≥ W_{1}(B_{2}). Let

β := \frac{W_{1}(B_{2}) − W_{2}(U + L)}{M} = k (2 + \frac{W_{2}(U + L)}{α})

and integrate (ii) from s ∈ [t−β, t] to t to get V (t)−V (s) ≤ M(t−s) or V (s) ≥ V (t)−M(t− s) ≥ W_{1}(B_{2})−Mβ = W_{2}(U +L). Thus, using (i) we obtain ∥x_{t_{n}}∥ ≥ W_{2}^{−1}(V (t_{n})) ≥ U +L for all t_{n} ∈ [t − β, t]. Define N to be the largest positive integer so that t_{N} ≤ t. Since ∥x_{t_{0}}∥ < B_{1} ≤ U + L there is an \overline{N} > 0 such that ∥x_{t_{n}}∥ ≥ U + L for n = \overline{N} + 1, ..., N, but ∥x_{t\overline{N}} ∥ < U + L. Note that from the above observation we know that

N − \overline{N} ≥ \frac{β}{k} − 2 = \frac{W_{2}(U + L)}{α}.  (2)

Let S_{n} = [t_{n−1}, t_{n}]. Since ∥x_{t_{n}}∥ ≥ U + L for n = \overline{N} + 1, ..., N there is a t′_{n} ∈ S_{n} such that |x(t′_{n})| ≥ U + L. There are two cases.

Case 1: If |x(s)| ≥ U for s ∈ S_{n} then integrating (iii) we obtain

V (t_{n}) − V (t_{n−1}) ≤ ∫_{t_{n−1}}^{t_{n}}[−a|x′(s)| − b]ds ≤ −bh < −α.

Case 2: If |x(s)| = U for some s ∈ [t_{n−1}, t_{n}] then without loss of generality we may assume that |x(u)| ≥ U for all u between s and t′_{n} and hence integrating (ii) and (iii) we have

V (t_{n}) − V (t_{n−1}) ≤ −| ∫_{s}^{t′_{n}}[a|x′(v)| + b]dv| + Mk ≤ −aL + Mk ≤ −α

again. In any case we find that V (t_{n}) − V (t_{n−1}) ≤ −α for n = \overline{N} + 1, ..., N. By the choice of \overline{N} we have

\begin{aligned}V (t_{N}) & = V (t\overline{N}) + ∑_{n=\overline{N}+1}^{N}(V (t_{n}) − V (t_{n−1})) ≤ V (t\overline{N}) − α(N − \overline{N}) \\ & ≤ W_{2}(∥x_{t\overline{N}} ∥) − α(N − \overline{N}) < W_{2}(U + L) − α(N − \overline{N}) ≤ 0 \end{aligned}

by (2), which is a contradiction to the fact that 0 ≤ V (t, x_{t}) and hence it proves that the solutions are uniformly bounded at t_{0}.

Consider now the uniform ultimate boundedness. Once again, we will prove everything for t_{0} only. The definition of L is slightly different now, we do not have B_{1} to ask for L ≥ B_{1}. So, choose L > 0 large enough to have −aL + Mk < −α for some α ∈ (0, bh). Let B > 0 be the B_{2} of the uniform boundedness (at t_{n} for any n) for B_{1} := U + L. Let B_{3} > 0 be given; we need to find a K > 0 such that if ϕ ∈ C, ∥ϕ∥ ≤ B_{3} and t ≥ t_{0} +K then |x(t)| = |x(t, t_{0}, ϕ)| < B. In order to do this we need to prove only that ∥x_{t_{n}}∥ < U + L for some t_{n} ∈ [t_{0}, t_{0} +K], since then from the uniform boundedness at t_{n} we obtain |x(s)| < B for all s ≥ t_{0} + K ≥ t_{n}. Define K := (W_{2}(B_{3})/α + 1)k and suppose for contradiction that ∥x_{t_{n}}∥ ≥ U + L for all t_{n} ∈ [t_{0}, t_{0} + K]. Let S_{n} := [t_{n−1}, t_{n}] for n = 1, 2, .... From here on, we do the same as we did in the proof of the uniform boundedness: find a t′_{n} ∈ S_{n} with |x(t′_{n})| ≥ U + L and using the two cases we prove that V (t_{n}) − V (t_{n−1}) ≤ −α for n = 1, 2, ... with S_{n} ⊂ [t_{0}, t_{0} + K]. The last estimate works out with a slight difference:

V (t_{N}) = V (t_{0}) + ∑_{n=1}^{N}(V (t_{n}) − V (t_{n−1})) ≤ W_{2}(B_{3}) − αN < 0

Uniform Boundedness and Uniform Ultimate Boundedness 8

if N > W_{2}(B_{3})/α, and we have this by our definition of K. This is a contradiction to the fact that 0 ≤ V (t_{N}) and hence it proves that the solutions are uniformly ultimately bounded at t_{0}.

Remark. Note that conditions (ii) and (iii) of the theorem are equivalent with the following: There are constants U, M > 0 such that (ii)’ V′(t, x_{t}) ≤ −W_{3}(|x(t)|) + M and (iii)’ V′(t, x_{t}) ≤ −W_{4}(|x′(t)|) + M for |x(t)| ≥ U, where W_{3}(U) > M and W_{4} is convex downward.

Proof: Assume (ii)’ and (iii)’ holds. Then for any α ∈ [0, 1] we have

V′(t, x_{t}) ≤ −αW_{3}(|x(t)|) − (1 − α)W_{4}(|x′(t)|) + M.

As W_{4} is convex downward there are constants c, d > 0 such that W_{4}(r) ≥ cr − d. Substituting this into the above estimate, assuming that |x(t)| ≥ U and reordering we have

\begin{aligned}V′(t, x_{t}) & ≤ (−αW_{3}(|x(t)|) + M) − (1 − α)(c|x′(t)| − d) \\ & = −(1 − α)c|x′(t)| + (−αW_{3}(U) + M + (1 − α)d), \end{aligned}

so taking α sufficiently close to 1 we have the desired (iii) estimate. Clearly (ii) holds.

To prove the opposite direction note that (iii) implies (iii)’. For (ii)’ define W_{3} by W_{3}(U) = 1, W_{3}(r) = M + b + 1 for r ≥ U + 1 and linear everywhere else. Then by (ii)

V′(t, x_{t}) ≤ M ≤ −W_{3}(|x(t)|) + (M + 1)

for |x(t)| ≤ U. Similarly, by (iii)

V′(t, x_{t}) ≤ −b ≤ −W_{3}(|x(t)|) + (M + 1)

for |x(t)| > U. Thus (ii)’ is satisfied by replacing U + 1 by U and M + 1 by M, and the proof is complete.

We now give an example for Theorem 3. Let a, b, c : R → R be continuous functions, h > 0, and consider

x′(t) = −a(t)x(t) + b(t)x(t − h) + c(t).  (3)

9 G. Makay

Suppose, that there are constants k > 0, B > 0 and α > 0 such that for p := 1 + α we have

(i) −a(t) + p|b(t + h)| ≤ min{−α, −αa(t)}, (ii) |c(t)| ≤ B and (iii) there is a sequence t_{n} ∈ R with h ≤ t_{n} − t_{n−1} ≤ k with ∫_{t_{n}−h}^{t_{n}} |b(s + h)|ds ≤ B. Then the solutions of (3) are uniformly bounded and uniformly ultimately bounded at t_{0}.

Proof: Define

V (t, x_{t}) = |x(t)| + p ∫_{t−h}^{t} |b(s + h)| · |x(s)|ds.

Clearly we have x(t) ≤ V (t, x_{t}) and

V (t_{n}, x_{t_{n}}) ≤ ∥x_{t_{n}}∥ + p ∫_{t_{n}−h}^{t_{n}} |b(s + h)|∥x_{t_{n}}∥ds ≤ (1 + pB)∥x_{t_{n}}∥.

Also, differentiating V we find

V′(t, x_{t}) ≤ −a(t)|x(t)| + |b(t)||x(t − h)| + |c(t)| + p|b(t + h)||x(t)| − p|b(t)||x(t − h)|
≤ [−a(t) + p|b(t + h)|]|x(t)| + (1 − p)|b(t)||x(t − h)| + B
≤ −α[a(t)|x(t)| + |b(t)||x(t − h)|] + B.

Using (i) from this last estimate we have V′ ≤ −α|x′(t)|+B and V′ ≤ −α|x(t)|+B, which is enough for Theorem 3 by the remark. Thus, the solutions of (3) are uniformly bounded and uniformly ultimately bounded at t_{0}.

This result implies the existence of a T -periodic solution if the functions a, b and c are T -periodic. There are other ways to prove the existence of a periodic solution for this special kind of differential equation. The author proved theorems in [8] that imply the existence of a periodic solution under a different set of conditions.

Consider now the functional differential equation

x′ = f(t, x_{t})  (4)

with infinite delay, i.e., let C be the Banach space of continuous functions ϕ : (−∞, 0] → R^{r} with ∥ϕ∥ := sup_{s∈(−∞,0]} |ϕ(s)| < ∞, f : R × C → R^{r} is continuous and locally Lipschitz in ϕ with this topology defined on C. Let the function g : (−∞, 0] → [1, ∞) be continuous

Uniform Boundedness and Uniform Ultimate Boundedness 10

and monotone decreasing with g(s) → ∞ as s → −∞ and g(0) = 1. We define the norm ∥ϕ∥_{g} := sup_{s≤0} |ϕ(s)|/g(s).

Definition. Solutions of (4) are uniformly bounded at t_{0}, if for all B_{1} > 0 there is a B_{2} > 0 such that ϕ ∈ C, ∥ϕ∥ < B_{1} and t ≥ t_{0} imply |x(t, t_{0}, ϕ)| < B_{2}.

Definition. Solutions of (4) are g-uniformly bounded at t_{0}, if for all B_{1} > 0 there is a B_{2} > 0 such that ϕ ∈ C, ∥ϕ∥_{g} < B_{1} and t ≥ t_{0} imply |x(t, t_{0}, ϕ)| < B_{2}.

Definition. Solutions of (4) are uniformly ultimately bounded with bound B at t_{0}, if for all B_{3} > 0 there is a K > 0 such that ϕ ∈ C, ∥ϕ∥ < B_{3} and t ≥ t_{0} + K imply |x(t, t_{0}, ϕ)| < B.

Definition. Solutions of (4) are g-uniformly ultimately bounded with bound B at t_{0}, if for all B_{3} > 0 there is a K > 0 such that ϕ ∈ C, ∥ϕ∥_{g} < B_{3} and t ≥ t_{0} + K imply |x(t, t_{0}, ϕ)| < B.

Since ∥ϕ∥_{g} ≤ ∥ϕ∥, g-uniform boundedness is a stronger condition than uniform boundedness and the same is true for uniform ultimate boundedness. The next theorem proves uniform boundedness and uniform ultimate boundedness for infinite delay equations. We suggest that the reader draw a picture for the proof of this theorem. Although the main idea is not very complicated, the interpretation is very technical.

Theorem 4: Suppose there are a continuous functional V : R × C → R, wedges W_{i} and constants M, U, a, b > 0 such that

(i) W_{1}(|ϕ(0)|) ≤ V (t, ϕ) ≤ W_{2}(∥ϕ∥_{g}) (ii) V′(t, x_{t}) ≤ −a|x′(t)| − b for |x(t)| ≥ U (iii) V′(t, x_{t}) ≤ M, where W_{1}(r) → ∞ as r → ∞. Also assume, that g(s) ≥ −μs for all s ≤ 0 and some μ > 0. Then the solutions of (4) are g-uniformly bounded and uniformly ultimately bounded.

Proof: First we prove the g-uniform boundedness. Let B_{1} > 0 be arbitrary and L ≥ B_{1} be large enough to have

μ (−a(U + L) + \frac{M}{μ} ) ≤ −2b

and

(−a(U + L) + \frac{M}{μ} ) ≤ −2aU.

11 G. Makay

The last estimation clearly implies aU + M/μ ≤ aL; we will need this later. Also define B_{2} by

\frac{b(W_{1}(B_{2}) − W_{2}(U + L))}{M} > W_{2}(U + L) + aL.

While these definitions are complicated, they will simplify later work. Suppose for contradiction that there is such a (t_{0}, ϕ) ∈ R × C such that ∥ϕ∥_{g} < B_{1} and |x(t, t_{0}, ϕ)| ≥ B_{2} for some t ≥ t_{0}. Using the very same idea as we used in Theorem 3, we can prove that if

β := \frac{W_{1}(B_{2}) − W_{2}(U + L)}{M}

then s ∈ (t − β, t] implies ∥x_{s}∥ ≥ ∥x_{s}∥_{g} ≥ W_{2}^{−1}(V (s)) > U + L. Let v_{0} ∈ [t_{0}, t] be the largest number such that ∥x_{v_{0}}∥_{g} ≤ U + L (since ∥x_{t_{0}}∥_{g} < B_{1} ≤ U + L we can find such a v_{0}). It is not difficult to see from the definition of v_{0} and the monotonicity of g that |x(v_{0})| = U + L. It is also clear by the above observation and the definition of B_{2} that

t − v_{0} ≥ β > \frac{W_{2}(U + L) + aL}{b}.  (5)

Let w_{0} = u_{0} = v_{0}, η_{0} = 1 and V_{0} := V (v_{0}) + aLη_{0}. We now define the sequences η_{n} and u_{n} ≤ v_{n} ≤ w_{n} by induction. The sequences will have the following property: either

(a) v_{n} = t and V (v_{n}) − V_{0} ≤ −b(v_{n} − v_{0}) or

(b) η_{n} ≥ 1, η_{n}(U + L) is the maximum of |x(s)| on the interval [u_{n}, v_{n}], V (v_{n}) − V_{0} ≤ −b(v_{n} − v_{0}) − aLη_{n} and sup_{s≤−(w_{n}−u_{n})} |x(w_{n} + s)|/g(s) ≤ U + L.

Clearly, (b) holds for n = 0. In our induction step we prove, that if (b) holds for η_{n}, u_{n}, v_{n} and w_{n}, then we can define η_{n+1}, u_{n+1}, v_{n+1} and w_{n+1} so that either (a) or (b) holds. To define u_{n+1}, v_{n+1}, w_{n+1} we consider two cases.

Case 1: If |x(s)| ≥ U for all s ∈ [v_{n}, t] then define v_{n+1} = t. Clearly, (ii) implies V (v_{n+1}) − V (v_{n}) ≤ −b(v_{n+1} − v_{n}) and hence (a) holds for v_{n+1}.

Case 2: If there is a u_{n+1} ∈ (v_{n}, t) such that |x(u_{n+1})| = U and |x(s)| ≥ U for s ∈ [v_{n}, u_{n+1}], then let η(U + L) be the maximum of |x(s)| on the interval [v_{n}, u_{n+1}] and w_{n+1} := u_{n+1} + (η + η_{n})/μ.

Case 2/a: If w_{n+1} ≥ t then we define v_{n+1} = t.

Uniform Boundedness and Uniform Ultimate Boundedness 12

Case 2/b: If w_{n+1} < t then ∥x_{w_{n+1}}∥_{g} > U + L. By the above definitions we obtain the followings:

sup_{s≤−(w_{n+1}−u_{n})} \frac{|x(w_{n+1} + s)|}{g(s)} ≤ sup_{s≤−(w_{n}−u_{n})} \frac{|x(w_{n} + s)|}{g(s)} ≤ U + L

sup_{−(w_{n+1}−u_{n})<s≤−(w_{n+1}−v_{n})} \frac{|x(w_{n+1} + s)|}{g(s)} ≤ \frac{η_{n}(U + L)}{η + η_{n}} ≤ U + L

sup_{−(w_{n+1}−v_{n})<s≤−(w_{n+1}−u_{n+1})} \frac{|x(w_{n+1} + s)|}{g(s)} ≤ \frac{η(U + L)}{η + η_{n}} ≤ U + L

because w_{n+1} − v_{n} ≥ w_{n+1} − u_{n+1} = (η + η_{n})/μ and hence if s ≤ −(w_{n+1} − v_{n}) ≤ −(w_{n+1} − u_{n+1}) = −(η + η_{n})/μ then g(s) ≥ η + η_{n}. Now putting these together we find

U + L < ∥x_{w_{n}}∥_{g} = sup_{−(w_{n+1}−u_{n+1})<s≤0} \frac{|x(w_{n+1} + s)|}{g(s)} ≤ sup_{−(w_{n+1}−u_{n+1})<s≤0} |x(w_{n+1} + s)|,

and hence we can find a v_{n+1} ∈ (u_{n+1}, w_{n+1}] such that |x(v_{n+1})| > U + L.

Now, in both Case 2/a and 2/b we estimate V (v_{n+1}) − V_{0}. In the following we will use several times that v_{n+1} − u_{n+1} ≤ w_{n+1} − u_{n+1} = (η + η_{n})/μ without even mentioning it. Define |x(s_{1})| = η_{n+1}(U + L) to be the maximum of |x(s)| on the interval [u_{n+1}, v_{n+1}], and let s_{2} ∈ [u_{n+1}, s_{1}] be the largest number with |x(s_{2})| ≤ U (in Case 2/a we do not have to do this, in this case we must drop the integral from s_{2} to s_{1} and −aLη_{n+1} from the following estimations). Integrating (ii) and (iii) we have

V (v_{n+1}) − V_{0} ≤ V (v_{n+1}) − V (v_{n}) + V (v_{n}) − V_{0}

≤ ∫_{v_{n}}^{u_{n+1}} −a|x′(s)| − bds + ∫_{s_{2}}^{s_{1}} −a|x′(s)| − bds + ∫_{u_{n+1}}^{v_{n+1}} Mds

− b(v_{n} − v_{0}) − aLη_{n}

≤ −a(η(U + L) − U) − b(u_{n+1} − v_{0}) + \frac{M(η + η_{n})}{μ} − aLη_{n} − aLη_{n+1}

≤ aU − b(u_{n+1} − v_{0}) + η (−a(U + L) + \frac{M}{μ} ) + (\frac{M}{μ} − aL) η_{n} − aLη_{n+1}

≤ aU + \frac{1}{2} (−a(U + L) + \frac{M}{μ} ) + (\frac{M}{μ} − aL) η_{n}

− b(u_{n+1} − v_{0}) + \frac{(v_{n+1} − u_{n+1})μ − η_{n}}{2} (−a(U + L) + \frac{M}{μ} ) − aLη_{n+1}

≤ −b(v_{n+1} − v_{0}) − aLη_{n+1} + \frac{1}{2} (\frac{M}{μ} + aU − aL) η_{n}

≤ −b(v_{n+1} − v_{0}) − aLη_{n+1}

13 G. Makay

by our definition of L. Therefore, we proved that in Case 2/a v_{n+1} satisfies (a), and in Case 2/b η_{n+1}, u_{n+1}, v_{n+1}, w_{n+1} satisfy (b).

This finishes our induction step. Continuing this process we either end up in Case 1 or Case 2/a, or we have sequences u_{n}, v_{n} and w_{n} for all n ≥ 0. Also, by (i) V_{0} ≤ W_{2}(∥x_{v_{0}}∥_{g}) + aLη_{0} ≤ W_{2}(U + L) + aL.

Case A: If we end up in Case 1 or in Case 2/a when we started our procedure from v_{n}, then v_{n+1} satisfies (a) and we have v_{n+1} = t and V (v_{n+1}) − V_{0} ≤ −b(v_{n+1} − v_{0}). Using these and (5) we obtain

V (t) ≤ V_{0} − b(t − v_{0}) ≤ W_{2}(U + L) + aL − bβ < 0,

which is a contradiction.

Case B: If we have the sequences u_{n}, v_{n} and w_{n} for all n ≥ 0, then v_{n} → v ≤ t as n → ∞ for some v. Then integrating (ii) and (iii) from v_{0} to v_{n} we have

\begin{aligned}V (v_{n}) − V (v_{0}) & ≤ M(v_{n} − v_{0}) − ∑_{i=0}^{n−1} ∫_{v_{i}}^{u_{i+1}}[a|x′(s)| + b]ds \\ & ≤ M(t − v_{0}) − ∑_{i=0}^{n−1} aL = M(t − v_{0}) − anL \end{aligned}

a contradiction for large n to 0 ≤ V (v_{n}).

This finishes the proof of the g-uniform boundedness.

For the uniform ultimate boundedness we define L > 0 as we did in the proof of g-uniform boundedness. Let B := B_{2} for B_{1} := U + L + 1 in the g-uniform boundedness. In order to prove uniform ultimate boundedness it is sufficient to prove that for all B_{3} > 0 there is a K > 0 such that for every (t_{0}, ϕ) with ∥ϕ∥ < B_{3} there is an s ∈ [t_{0}, t_{0} + K] with ∥x_{s}(·, t_{0}, ϕ)∥_{g} ≤ U + L; then g-uniform boundedness implies the necessary inequality. Let B_{3} > 0 be arbitrary, D := B_{3}/(μ(U + L)), w_{0} := t_{0} + D and η_{0}(U + L) to be the maximum of |x(s)| on the interval (−∞, w_{0}]. As x is bounded by B_{3} on (−∞, t_{0}] and V (s) ≤ V (t_{0}) + M(s − t_{0}) ≤ W_{2}(B_{3}) + MD, we have |x(s)| ≤ max{B_{3}, W_{1}^{−1}(W_{2}(B_{3}) + MD)\} =: \overline{η}(U + L) for s ≤ w_{0} and hence η_{0} ≤ \overline{η}, a bound independent of t_{0} and ϕ. Define K := D + (W_{2}(B_{3}) + DM + aL\overline{η})/b + 1 and t := t_{0} + K. Suppose for contradiction, that there is a (t_{0}, ϕ) such that ∥x_{s}∥_{g} = ∥x_{s}(·, t_{0}, ϕ)∥_{g} > U + L for all s ∈ [t_{0}, t_{0} + K]. Then using the below estimation on g we find

U + L < ∥x_{w_{0}}∥_{g} = sup_{s≤0} \frac{|x(w_{0} + s)|}{g(s)}

≤ max \{ sup_{s≤−D} \frac{|x(w_{0} + s)|}{g(s)}, sup_{−D≤s≤0} \frac{|x(w_{0} + s)|}{g(s)}\}

≤ max \{\frac{B_{3}}{μD} , sup_{−D≤s≤0} \frac{|x(w_{0} + s)|}{g(s)}\}

≤ max {U + L, sup_{−D≤s≤0} |x(w_{0} + s)|} ,

and hence we find a v_{0} ∈ [t_{0}, w_{0}] with |x(v_{0})| > U + L. Let u_{0} = −∞. From here on the proof is exactly the same as we did before. Note, that V (t_{0}) ≤ W_{2}(∥ϕ∥_{g}) ≤ W_{2}(∥ϕ∥) ≤ W_{2}(B_{3}) and hence V (v_{0}) ≤ W_{2}(B_{3}) + (v_{0} − t_{0})M ≤ W_{2}(B_{3}) + DM. We have Cases A and B again:

Case A: If v_{n+1} = t, then from v_{0} ≤ t_{0} + D we have t − v_{0} ≥ K − D > (W_{2}(B_{3}) + DM + aL\overline{η})/b and hence V (v_{n+1}) − V_{0} ≤ −b(v_{n+1} − v_{0}) implies

V (t) ≤ V_{0} − b(t − v_{0}) < W_{2}(B_{3}) + DM + aLη_{0} − b\frac{W_{2}(B_{3}) + DM + aL\overline{η}}{b} ≤ 0

a contradiction.

Case B: We get a contradiction as before.

This proves the uniform ultimate boundedness and the proof is complete.

Note that with a little more care one can prove the exact counterpart of Theorem 3 for infinite delay equations too. The theorem would be the following.

Theorem 5: Suppose there are a continuous functional V : R × C → R, wedges W_{i}, constants M, U, a, b, k > 0 and a sequence t_{n} ∈ R with 1 ≤ t_{n} − t_{n−1} ≤ k such that

(i) W_{1}(|ϕ(0)|) ≤ V (t, ϕ) and V (t_{n}, ϕ) ≤ W_{2}(∥ϕ∥_{g}) (ii) V′(t, x_{t}) ≤ −a|x′(t)| − b for |x(t)| ≥ U (iii) V′(t, x_{t}) ≤ M, where W_{1}(r) → ∞ as r → ∞. Also assume, that g(s) ≥ −μs for all s ≤ 0 and some μ > 0. Then the solutions of (4) are g-uniformly bounded and uniformly ultimately bounded at t_{n}.

15 G. Makay

Also, B_{2} of the uniform boundedness and B, K of uniform ultimate boundedness does not depend on n.

Proof: We only mention the changes one needs to make on the proof of Theorem 4. We need to modify the definition of w_{n+1} (starting with w_{0}) so that w_{n+1} is one of the t_{i}’s in the interval of length k with the left-end point being the previous definition of w_{n+1}, and hence we have V (w_{n+1}) ≤ W_{2}(∥x_{w_{n+1}}∥_{g}). Also, because we have w_{n+1} ≤ u_{n+1} + (η + η_{n})/μ + k only, we need to do the estimation on V (v_{n+1}) − V_{0} more carefully, we will have some terms containing k too. To get the same estimation as before, instead of cutting the term η(−a(U +L)+M/μ) into two equal pieces we need to use an α ∈ (0, 1) to cut it. According to this α we need to modify the definition of L too.

References:

[1] T. A. Burton, Stability and periodic solutions of ordinary and functional differential equations, Academic Press, Orlando, Florida, (1985).

[2] Burton, T. A. and Zhang, B., Uniform ultimate boundedness and periodicity in functional differential equations, Tohoku Math. J. 42 (1990), 93-100.

[3] He, M., Huang, Q. and Wang, K., Phase spaces C_{g}, C_{h} and g-uniform boundedness of FDE(ID), J. Math. Anal. Appl. 138 (1989), 473-490.

[4] Hering, R. H., Boundedness and stability in functional differential equations, dissertation at SIUC (1988).

[5] Hering, R. H., Boundedness and periodic solutions in infinite delay systems, J. Math. Anal. Appl. 163 (1992), 521-535.

[6] Kato, J., An autonomous system whose solutions are uniformly ultimately bounded but not uniformly bounded, Tohoku Math. J. 32 (1980), 499-504.

[7] G. Makay, Periodic solutions of dissipative functional differential equations, Tohoku Math. J. 46 (1994), 417-426.

[8] G. Makay, Periodic solutions of linear functional differential equations, J. Differential and Integral

Equations 8 (1995), 2177-2187.

Department of Mathematics  Bolyai Institute
Southern Illinois Univeristy  Aradi vértanuk tere 1
Carbondale, IL 6290-4408  Szeged, H-6720
U.S.A.  Hungary