The characteristic equation’s eigenvalues including negative real components are a necessary and sufficient condition for the system to remain stable. Following is the Jacobian matrix for the new system (1) up to E = (0, 0, 0, 0, 0, 0):

J = [ − 10 10 0 1 0 0 28 − 1 0 0 1 0 0 0 − 2.6 0 0 0 0 0 0 2 0 0 0 − 8.4 0 0 0 0 0 1 0 0 0 − 1 ] (3)

The characteristic equation is:

λ 6 + 12.5 λ 5 − 248.6 λ 4 − 212.76 λ 3 + 1387.12 λ 2 + 50.4 λ − 436.8 = 0 , (4)

Roots of the characteristic equation:

λ 1 = − 22.823 , λ 2 = − 2.617 , λ 3 = − 0.574 , λ 4 = 0.581 , λ 5 = 1.860 , λ 6 = 10.940

Thus, we conclude that it is unstable system.

A system meets the Routh requirement for stability (all poles in the half-loop level), if and only if the entries in the Routh array’s first column have values that are entirely positive. The number of sign changes in the first column multiplied by the sum of the non-OLHP columns [25]. Regarding the Roth stability test, see Table 1.

a 0 = − 436.8 , a 1 = 50.4 , a 2 = 1387.12 , a 3 = − 212.76 , a 4 = − 248.6 , a 5 = 12.5 , a 6 = 1 ,

b 0 = a 5 a 0 − a 6 a 7 a 5 = − 436.8 ,     c 1 = b 4 a 1 − a 5 b 0 b 4 = 26.822 ,

b 2 = a 5 a 2 − a 6 a 1 a 5 = 1383.088 ,     c 3 = b 4 a 3 − a 5 b 2 b 4 = − 138.104 ,

b 4 = a 5 a 4 − a 6 a 3 a 5 = − 231.579 ,     e 1 = c 3 d 0 − c 1 d 2 d 2 = 18.259 ,

d 0 = b 4 − b 0 c 3 c 3 = 436.8 ,     d 2 = b 4 c 1 − b 2 c 3 c 3 = − 1338.111.

The system is unstable because the first column has four negative elements.

Determinants generated from the coefficients of the characteristic equation are used to implement this criterion. System (1) is stable if the tiny minors of its square matrix J are all positive; if not, it is unstable [25] [26].

If n = 6 ,

From Equation (3):

Δ 1 = a n − 1 = a 5 = 12.5 > 0 ,

Δ 2 = | a n − 1 a n − 3 a n a n − 2 | = | a 5 a 3 a 6 a 4 | = | 12.5 − 212.76 1 − 248.6 | = − 2894.74 < 0 ,

Δ 3 = | a n − 1 a n − 3 a n − 5 a n a n − 2 a n − 4 0 a n − 1 a n − 3 | = | 12.5 − 212.76 50.4 1 − 248.6 1387.12 0 12.5 − 212.76 | = 39977.383 > 0 ,

Δ 4 = − 56781.350 < 0 ,

Δ 5 = 79321.51 > 0 ,

Δ 6 = 103579.12 < 0.

System (1) is unstable because some of the values of the determinants are less than zero.

Where we assume the Lyapunov function is:

V ( x 1 , x 2 , x 3 , x 4 , x 5 , x 6 ) = 1 / 2 ( x 1 2 + x 2 2 x 3 2 + x 4 2 + x 5 2 + x 6 2 )

V ˙ ( x 1 , x 2 , x 3 , x 4 , x 5 , x 6 ) = x 1 x ˙ 1 + x 2 x ˙ 2 + x 3 x ˙ 3 + x 5 x ˙ 5 + x 6 x ˙ 6

V ˙ = 38 x 1 x 2 − 10 x 1 2 + x 1 x 4 − x 2 2 − 7.4 x 2 x 5 − 2.6 x 3 2 + 2 x 4 2 − x 1 x 3 x 4 + x 2 x 6 2 − x 6 2 . (5)

Since V ˙ ( x 1 , x 2 , x 3 , x 4 , x 5 , x 6 ) > 0 , as a result, New system (1) is unstable.

By creating a continuous fraction from the odd and even parts of the equation, the characteristic equation of a continuous system is subjected to this condition [25]. The distinguishing equation:

λ 6 + 12.5 λ 5 − 248.6 λ 4 − 212.76 λ 3 + 1387.12 λ 2 + 50.4 λ − 436.8 = 0 ,

By taking the even terms and then the odd terms, respectively, we have:

Q 1 ( λ ) = λ 6 − 248.6 λ 4 + 1387.12 λ 2 − 436.8 , (6)

Q 2 ( λ ) = 12.5 λ 5 − 212.76 λ 3 + 50.4 λ , (7)

After dividing the even terms by the odd terms and using algebraic steps, we get the following results:

h 1 = 0.08 , h 2 = − 0.053 , h 3 = 0.809 , h 4 = − 0.21 , h 5 = − 21.108 , h 6 = 0.147.

Since some values of h are negative, the equation of the system has some positive real roots, so system (1) is chaotic.

Suppose that

f 1 = d x 1 d t , f 2 = d x 2 d t , f 3 = d x 3 d t , f 4 = d x 4 d t , f 5 = d x 5 d t and f 6 = d x 6 d t .

The obtained vector field,

V ( x ˙ 1 , x ˙ 2 , x ˙ 3 , x ˙ 3 , x ˙ 4 , x ˙ 5 , x ˙ 6 ) T = ( f 1 , f 2 , f 3 , f 4 , f 5 , f 6 ) T

∇ ⋅ ( x ˙ 1 , x ˙ 2 , x ˙ 3 , x ˙ 3 , x ˙ 4 , x ˙ 5 , x ˙ 6 ) T = ∂ f 1 ∂ x 1 + ∂ f 2 ∂ x 2 + ∂ f 3 ∂ x 3 + ∂ f 4 ∂ x 4 + ∂ f 5 ∂ x 5 + ∂ f 6 ∂ x 6 = − ( a + 1 + b − d + h ) = f .

Note that, f = − ( a + 1 + b − d + h ) = − 12.6 , for all ( a , b , d , h ) values that are positive and greater than zero, the system (1) is dissipative

The exponential rate is:

d V d t = f V ⇒ V ( t ) = V 0 e f t = V 0 e − 12.6 t

By flowing into ( V 0 e − 12.6 ), the volume element ( V 0 ) from the previous equation is condensed at the time (t).

The fifth- and sixth-order Runge-Kutta method is used to solve the system (1). Initial values included

x | x 1 ( 0 ) , x 2 ( 0 ) , x 3 ( 0 ) , x 4 ( 0 ) , x 5 ( 0 ) , x 6 ( 0 ) = [ 3 , 2 , 0.5 , 1 , 2.5 , 3.5 ]

The waveform exhibits aperiodic structure, the primary defining feature of chaotic systems. x 1 ( t ) , x 2 ( t ) , x 3 ( t ) , x 4 ( t ) , x 5 ( t ) and x 6 ( t ) for system (1) (Figure 1).

Figure 2 and Figure 3 in this paragraph depict the chaotic strange attractor for the system (1) in ( x 1 , x 2 , ⋯ , x 6 ) space and the chaotic strange attractor for the system (1) in ( x 1 , x 6 ) the plane, respectively.

Since the orbit in each graph looks to be dense, the new system exhibits a chaotic attractor.

The average exponential rates of almost divergent trajectories in phase space are frequently referred to as the Lyapunov exponent. The new system is regarded as chaotic if it has at least one positive Lyapunov exponent. Values of the Lyapunov exponent are:

L 1 = 0.524 , L 2 = − 0.279 , L 3 = 2.987 , L 4 = − 5.435 , L 5 = − 9.21 , L 6 = − 12.889.

As a result, the system’s “Kaplan-Yorke dimension” or Lyapunov dimension is as follows:

D L = 3 + L 1 + L 2 + L 3 | L 4 | = 3.59466

Figure 4 show that system (1) is very Chaotic.

To stabilize a chaotic system (1) use the sufficiency control law generalized with an unknown parameter c as follows:

x ˙ 1 = 10 ( x 2 − x 1 ) + x 4 + u 1 x ˙ 2 = c x 1 − x 2 − x 1 x 3 + x 5 + u 2 x ˙ 3 = − 2.6 x 3 + x 1 x 2 + u 3 x ˙ 4 = 2 x 4 − x 1 x 3 + u 4 x ˙ 5 = − 8.4 x 2 + u 5 x ˙ 6 = x 2 − x 6 + u 6 (8)

where [ u 1 , u 2 , u 3 , u 4 , u 5 , u 6 ] T are feedback controllers.

We now consider the following adaptive control procedures to make sure the managed system (8) converges asymptotically to the origin.

u 1 = − 10 ( x 2 − x 1 ) + x 4 − μ 1 x 1 u 2 = − c ^ x 1 + x 2 + x 1 x 3 − x 5 − μ 2 x 2 u 3 = 2.6 x 3 − x 1 x 2 − μ 3 x 3 u 4 = − 2 x 4 + x 1 x 3 − μ 4 x 4 u 5 = 8.4 x 2 − μ 5 x 5 u 6 = − x 2 + x 6 − μ 6 x 6 (9)

where μ 1 , μ 2 , μ 3 , μ 4 , μ 5 , μ 6 are constants, c ^ is an estimator of the parameter c.

Substituting (9) into (8), we get:

x ˙ 1 = − μ 1 x 1 x ˙ 2 = ( c − c ^ ) x 1 − μ 2 x 2 x ˙ 3 = − μ 3 x 3 x ˙ 4 = − μ 4 x 4 x ˙ 5 = − μ 5 x 5 x ˙ 6 = − μ 6 x 6 (10)

Let the error of estimating parameter is:

e c = c − c ^ (11)

Using (11), system (10) can be written as:

x ˙ 1 = − μ 1 x 1 x ˙ 2 = e c x 1 − μ 2 x 2 x ˙ 3 = − μ 3 x 3 x ˙ 4 = − μ 4 x 4 x ˙ 5 = − μ 5 x 5 x ˙ 6 = − μ 6 x 6 (12)

The parameter estimate c ^ is changed using the Lyapunov method of obtaining the updated law. It is thought that the quadratic Lyapunov function:

V ( x 1 , x 2 , x 3 , x 4 , x 5 , x 6 , e c ) = 1 2 ( x 1 2 + x 2 2 + x 3 2 + x 4 2 + x 5 2 + x 6 2 + e c 2 ) , (13)

Which definite, positive-in ℝ 7 .

Also

e ˙ c = − c ^ ˙ . (14)

Differentiate V & substituting (11) and (13), we get:

V ˙ ˙ = − μ 1 x 1 2 − μ 2 x 2 2 − μ 3 x 3 2 − μ 2 x 4 2 − μ 5 x 5 2 − μ 6 x 6 2 + e c ( x 1 x 2 − c ^ ˙ ) .

Assume that:

c ^ ˙ = x 1 x 2 + μ 7 e c . (15)

where μ 7 is greater than zero.

Substitute (15) into V ˙ ˙ , we get:

V ˙ ˙ = − μ 1 x 1 2 − μ 2 x 2 2 − μ 3 x 3 2 − μ 4 x 4 2 − μ 5 x 5 2 − μ 6 x 6 2 − μ 7 e c 2 . (16)

Which is negative-definite on ℝ 7 .

The outcome is as follows because of Lyapunov stability, Eigenvalues, and the Routh array criteria.

Proposition 1. By adaptive control (9), where c ^ ˙ = x 1 x 2 + μ 7 e c and μ 1 , μ 2 , − μ 3 , − μ 4 , − μ 5 , μ 7 are positive constants, The chaotic system (8) is stabilized for x ( 0 ) ∈ ℝ 6 .

The controlled extremely chaotic system (10) was simulated using

x | x 1 ( 0 ) , x 2 ( 0 ) , x 3 ( 0 ) , x 4 ( 0 ) , x 5 ( 0 ) , x 6 ( 0 ) = [ − 4 , 5 , 2 , 1 , − 2.5 , 3 ]

μ 1 , μ 2 , μ 3 , μ 4 , μ 5 , μ 6 = [ 10 , 6 , 4 , 7 , 5 , 8 , 3 ] and e c = 19 .

The new system (1)’s-controlled state trajectories are displayed in Figure 5.