Sliding Intermittent Control for BAM Neural Networks with Delays

Remark 1. The above conditions are general in the literature to study the existence and uniqueness of the equilibrium for the delayed BAM neural networks (1) without assuming the activation functions to be monotonic or differentiable [27–29].

Letting  (w^*, z^*)  with$$ and $$be an equilibrium of the system (1) and denoting$$, the equilibrium of the network (1) can be transformed to the origin of the following system:

where$$and$$.

Remark 2. The sliding intermittent control idea is illustrated in Figure 1 with the adjustable parameter  θ  (i.e., the control width  θ  shown in Figure 1). The closed-loop system (4) can be the continuous controlled neural networks (θ = 1), the impulsive controlled neural networks (θ = 0), or the hybrid controlled neural networks (0 < θ < 1).

The system (4) is supplemented with initial function given by  x_i(s) = φ_i(s),  y_j(s) = ψ_j(s),   − τ^* ≤ s ≤ 0, where  φ(s) = (φ₁(s), φ₂(s), …, φ_n(s)) ^T ∈ C([−τ^*, 0], ℝⁿ),  ψ(s) = (ψ₁(s), ψ₂(s), …, ψ_m(s)) ^T ∈ C([−τ^*, 0], ℝ^m), and  C([−τ^*, 0], ℝⁿ)  represents the set of all  n-dimensional continuous functions defined on the interval  [−τ^*, 0]. Obviously, the solution of (4) is piecewise left-hand continuous with possible discontinuity at  t = t_k  for  k ∈ ℕ⁺.

Definition 3. The system (4) is said to be globally exponentially stable if there exist constants  α > 0, β > 0  such that for all  t > 0,

holds for all  φ(·) ∈ C([−τ^*, 0], ℝⁿ)  and  ψ(·) ∈ C([−τ^*, 0], ℝ^m).

Lemma 4 (see [30].)Let  V(·):[t₀ − τ, ∞)→[0, ∞)  be a continuous function such that

is satisfied for  t ≥ t₀. If  a > b > 0, then where$$,  λ > 0  is the unique positive real root of the equation   − a + λ + be^λτ = 0.

Lemma 5 (see [31].)Let  q ≥ 0,  τ > 0,  μ_k > 0  (k = 1,2, …), and  p  be constants, and assume that  V(t)  is a piecewise continuous nonnegative function satisfying

If there exists constant  β  such that hold for  k = 1,2, …, then where$$,$$, and  λ  is the unique positive root of the equation  λ + p + dqe^λτ + β = 0.

Theorem 6. Assume the upper bound delay  τ^* < min  {θT, (1 − θ)T}  and the external imposed impulsive strengths satisfy$$, $$. Under the sliding intermittent control, the closed-loop control system (4) is globally exponentially stable if there exist constants  β  and$$, $$such that the following conditions hold:

(i)

(ii)

(iii)

where$$;$$;$$;  ρ = max _i,j {η − a_i, η − b_j};$$;$$is used to denote the last impulsive moment on the interval  [(l + θ)T, (l + 1)T)  with$$;  γ = τ^*/T;  0 < η ≤ η^*;$$with functions  F_i(·), G_j(·)  defined as

and  λ₁, λ₂  are, respectively, the unique positive real root of the equations$$and$$.

More specifically, we have the following inequality:

with$$.

Proof. For the functions  F_i(·)  and  G_j(·)  defined in (16), from the condition (13), it is clear that

Since  F_i(·)  and  G_j(·)  are continuous on  [0, ∞) and $$,$$, there must exist constants$$and$$such that$$ and$$. By setting$$, one obtains that, for any  0 < η ≤ η^*, Now consider the Lyapunov function defined as follows: where  u_i(t) = e^ηt | x_i(t) |   and  v_j(t) = e^ηt | y_j(t)|. Obviously,  V(t)  is a positive definite function for  t ≥ 0.

(1) When$$, one is easy to have

Calculating the upper right Dini derivative of  V(t)  along the solutions of network (4), from the above inequality, we get where$$ and$$. By Lemma 4, one has where  λ₁  is the unique positive real root of the equation$$.

(2) When  t ∈ [(l + θ)T, (l + 1)T) and  t ≠ t_k, $$, k ∈ ℕ⁺, it is easy to have

Without loss of generality, suppose  t ∈ (t_k−1, t_k], k ≤ i_l, where$$is assumed to be the last impulsive moment on the interval  [(l + θ)T, (l + 1)T). It follows from the inequality (24) that where  ρ = max _i,j {η − a_i, η − b_j};$$.

When  t ∈ [(l + θ)T, (l + 1)T)  and  t = t_k,

Considering the condition that$$, $$, we have  μ_k > 0 and which infers that From condition (14) and Lemma 5, one has where$$ and λ₂  is the unique positive real root of the equation$$.

(3) Now, we are ready to estimate  V(t)  based on the inequalities (23) and (29) with the method of induction.

•  

  When  t ∈ [0, θT), one obtains

  $$ ()

•  

  When  t ∈ [θT, T), one can derive

  $$ ()

•  

  When  t ∈ [T, (1 + θ)T), we have

  $$ ()

•  

  When  t ∈ [(1 + θ)T, 2T), one can derive

  $$ ()

By induction, one can derive the following estimation of  V(t)  for any integer$$:

•  

  when  t ∈ [lT, (l + θ)T),

  $$ ()

•  

  and when  t ∈ [(l + θ)T, (l + 1)T),

  $$ ()

By setting  γ = τ^*/T  and substituting it to the above two inequalities, one has, for  t ∈ [lT, (l + θ)T),

and, for  t ∈ [(l + θ)T, (l + 1)T), Therefore, we have which means that, for any  t > 0, where$$. And the conclusion that the origin of network (4) is exponentially stable. The proof is complete.

Remark 7. In the above Theorem 6, the control width  θ ∈ (0,1)  which does not include the boundary cases. If the parameter  θ → 0, the controlled system approximates to impulsive neural networks. If the parameter  θ → 1, the controlled system approximates to continuous neural networks. And the controlled system can be handled, respectively, by the methods of the impulsive system and the continuous system. In order to avoid such boundary cases, usually we can take  θ = 0.5; that is, the periodically intermittent controller and the impulsive controller play a significant role in the process of control; this results a switched neural networks.

Remark 8. Most of the literatures [28, 32] concerning the global exponential stability of the delayed BAM neural networks with impulses have focused on the stable system. Namely, without the impulsive disturbance, the original neural networks are stable, and under the impulsive disturbance the system can still be kept stable with particular conditions. While, in this article, the impulses can be viewed as either control input or impulsive disturbance, at the same time, the original system is not required to be stable initially.

If adjustable parameter  θ = 1, the controlled system turns out to be the following model; that is, the original unstable system (3) is controlled with the continuous feedback controller. Under such case, the conditions in the following proposition will guarantee the closed-loop system to be globally exponentially stable:

Proposition 9. Under the continuous feedback control scheme, the origin of the closed-loop control system (40) is globally exponentially stable if there exist constants$$and$$such that

Furthermore, one has the following inequality: where the parameters  η  and$$are consistent with the ones in Theorem 6.

Proof. Consider the Lyapunov function defined as follows:

The upper right Dini derivative of  W(t)  with respect to time  t  along the solutions of the network (40) can be calculated as follows: which means that  W(t) ≤ W(0). Hence we have $$, and this completes the proof.

Lemma 10 (see [23].)Let  V(·):  [t₀ − τ, ∞)→[0, ∞)  be a continuous function such that

is satisfied for  t ≥ t₀. If  a > 0,  b > 0, then

Proposition 11. Assuming the upper bound delay  τ^* < min  {θT, (1 − θ)T}, under the periodically intermittent control, the closed-loop control system (45) is globally exponentially stable if the control gains$$ and$$satisfy the following conditions:

• (i)
  $$ ()

• (ii)
  $$ ()

Moreover, we have where$$and  κ  is any positive constant; $$, and the parameters$$, $$, η, λ₁, and γ  are consistent with those in Theorem 6.

Proof. Considering the same Lyapunov function as that in Theorem 6, by similar analytical technique, one can get the following results on the control period and the control width as follows.

• (1)

  When$$, we have$$. By Lemma 4, one obtains that, for any  t ∈ [lT, (l + θ)T), $$.

• (2)

  When$$, we have$$. By Lemma 10, it is derived that, for any  t ∈ [(l + θ)T, $$.

From the above inequality relationships, by similar estimation procedure, we can get the following conclusion:

By the notational expression  τ^* = γT, one can further obtain Hence it follows that, for any  t > 0, and this completes the proof.

Proposition 12. Assume the external imposed impulsive strengths$$. The origin of the closed-loop control system (54) is globally exponentially stable if there exists constant  β  such that, for  k = 1,2, …, the following conditions hold:

More specifically, we have where the parameters  μ_k, ρ, d, $$, and  η  are consistent with those in Theorem 6.

Proof. Considering the same Lyapunov function as that in Theorem 6, by similar analytical technique, one can get the following results on the impulsive interval and the impulse moments as follows.

• (1)

  When  t ∈ (t_k−1, t_k], the upper right Dini derivative of  V(t)  along the solution of (54) is depicted as$$.

• (2)

  When  t = t_k,   k ∈ ℕ⁺,$$.

From Lemma 5 and the conditions in (55), it follows that

which means and the proof is completed.

Definition 13 ([21] average impulsive interval). The average impulsive interval of the impulsive sequence$$is equal to  T_a  if there exist positive integer  N₀  and positive number  T_a  such that

where  N(κ₂, κ₁)  denotes the number of impulsive times of the impulsive sequence$$on the interval  (κ₁, κ₂).

Lemma 14 (see Lakshmikantham et al., Theorem  1.4.1, [33], [34].)Let  PC(ℝ₊, ℝ)  (PC¹(ℝ₊, ℝ)) denote the set of piecewise continuous (piecewise continuously differentiable) functions with first kind of discontinuities at  t_k, k = 1,2, …, from  ℝ₊  to  ℝ. If the following conditions hold,

where  p(t), q(t) ∈ PC(ℝ₊, ℝ),  d_k ≥ 0  and  b_k  are real constants, then

Lemma 15. Let  q ≥ 0,  τ > 0,  0 < μ < 1, and  p  be constants, and assume that  V(t)  is a piecewise continuous nonnegative function satisfying

and the average impulsive interval of the impulsive sequence$$is equal to  T_a. If the following inequality holds, then one has where$$,  λ  is the unique positive root of the equation  λ + p + (ln μ/T_a) + $$  =  0.

Proof. By Lemma 14 and the definition of average impulsive interval, it follows from (63) that, for  t ≥ t₀,

Denote$$. Since$$, and  lim _λ→∞ϕ(λ) = +∞  and$$, one knows that  ϕ(λ) = 0  has a unique positive root.

Next, it is claimed that, for all  t > t₀,

When  t ∈ [t₀ − τ, t₀],

Supposing (67) is not always true for  t > t₀, there must exist one time point  t^* > t₀  such that

From inequalities (66) and (69), one can obtain that Considering the equality$$, one has which contradicts with the first inequality of (69). Therefore, the inequality (67) holds, and this completes the proof.

Proposition 16. Assume the external imposed impulsive strengths$$, $$, and the average impulsive interval of the impulsive sequence$$is equal to  T_a. The impulsive system (59) is globally exponentially stable if the following condition holds:

Moreover, we have where  μ = sup _k {μ_k} ∈ (0,1), and  μ_k, ρ, and $$are defined as in Theorem 6 and  λ  is the unique positive root of the equation$$.

Proof. Considering the same Lyapunov function as that in Theorem 6, the following results on the impulsive interval and the impulse moments can be obtained.

• (1)

  When$$, the upper right Dini derivative of  V(t)  along the solutions of (59) is depicted as$$.

• (2)

  When  t = t_k, k ∈ ℕ⁺,$$.

From the above two inequality relationships, and Lemma 15, one has

where  λ  is the unique positive root of the equation$$. The proof is completed.

Remark 17. It should be pointed out that, in the preceding control width of the control period, other kinds of continuous controllers can also be used to achieve the same performance. For example, in [35], the adaptive control scheme has been employed in the control width instead of the continuous state feedback, where the adjusting gains can be designed based on different norms. We can borrow such an idea to the sliding intermittent control design. Moreover, the sliding width parameter can be  {θ_i}  rather than the fixed width  θ, and the period can be  {T_i}  rather than constant  T. By doing so, we might obtain more general conditions. On the other hand, the phenomena of stochastic nonlinearities are extremely ubiquitous in practical controlled systems [36–39]; hence it is more reasonable to consider neural networks with random nonlinearities, and this will be our future works.

Example 1. Considering the following extensively studied BAM neural system,

with  a = 1.922,  p = 9.8501,  b = 1.1631,  q = 8.2311,  τ = σ = 3, and  f(x) = g(x) = 1/(1 + e^−x) − 1/2. Obviously,  L^f = L^g = 0.25. The initial condition is given as  x(t) = −0.43, y(t) = 0.42, t ∈ [−3,0]. With the above system parameters, the phase diagram of system (75) is given in Figure 2. Obviously, the origin of system (75) is not stable. We will design the sliding intermittent controller to stabilize it.

Remark 18. From the above verifying process, it can be found that the sliding intermittent control is much better than the pure periodically intermittent control. More specifically, in the periodically intermittent control, the control period  T = 100  and the control width  θ = 0.965, while in the sliding intermittent control, the control period  T = 10  and the control width  θ = 0.5. As for the impulsive control, the full impulsive control is better than the semi-impulsive control in that the earlier converges faster. When dealing with the nonuniformly distributed impulsive sequence, the result derived in Proposition 16 is less conservative.

Example 2. Consider the following unstable delayed BAM neural network, and we will show that the sliding intermittent control benefits the stabilization of the unstable system:

where  A = diag {3.1220,2.3156,2.2683},  B = diag {2.9631,2.3456,2.6341,3.0726},