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This paper investigates the maximum interval of stability and convergence of solution of a forced Mathieu’s equation, using a combination of Frobenius method and Eigenvalue approach. The results indicated that the equilibrium point was found to be unstable and maximum bounds were found on the derivative of the restoring force showing sharp condition for the existence of periodic solution. Furthermore, the solution to Mathieu’s equation converges which extends and improves some results in literature.

Consider a harmonically forced Mathieu’s equation defined by

d 2 x d t 2 + ( w + ε cos t ) x = f cos λ t (1)

For small | ε | , this equation describes a simple harmonic oscillator whose frequency is a periodic function of time with the boundary condition as;

x ( 0 ) = x ( 2 π )

x ˙ ( 0 ) = x ˙ ( 2 π ) (2)

where d 2 x d t 2 is the second derivative with respect to time, f is the amplitude of a

periodic driving force, w and ε are the Mathieu’s parameters and λ is the angular frequency of the periodic driving force.

Mathieu’s equation is a special case of a linear second order homogenous equation [

Stability is an important concept in linear and nonlinear analysis. For instance, roughly speaking, a physical system is stable if small changes at sometimes cause only a small change in the behavior of the system in future [

Motivated by the above literature and ongoing research in this direction, the objectives of this paper are to investigate the maximum interval of stability and convergence of solution of forced Mathieu’s equation. We further prove that the solution converges in that interval of interest and that if all solutions are bounded, then the corresponding point in the w and ε parameter plane is said to be stable.

Definition 2.1. Frobenius method which was named after Ferdinard Geog Frobenius is a method for finding an infinite series solution for a second-order ordinary differential equation of the form;

x 2 y ″ + p ( x ) y ′ + q ( x ) y = 0 (3)

with y ′ = d y d x and y ″ = d 2 y d x 2

In the vicinity of the regular singular point x = 0 , we can divide (3) by x 2 to obtain a differential equation of the form;

y ′ ′ + p ( x ) x 2 y ′ + q ( x ) x 2 y = 0 (4)

which will not be solvable with regular power series method if either p ( x ) x 2 or q ( x ) x 2 are not analytic at x = 0 . The Frobenius method enables us to make use

of powers series solution to solve a different equation, given that p ( x ) and q ( x ) are themselves analytic at 0 or being analytic elsewhere. In this case both limit exist at 0 and are finite [

Definition 2.2. Stability is a qualitative property of behavior of the solutions of differential equations by which given a reference solution x * ( t , t 0 * , x 0 * ) of x ˙ = x ( t , t 0 * , x 0 * ) ∈ ℝ n , t ≥ t 0 * , t 0 * ≥ 0 , any other solution x ( t , t 0 , x 0 ) starting close to x * ( t , t 0 * , x 0 * ) i.e. t 0 * ~ t 0 and x 0 * ~ x 0 remains close to x * ( t , t 0 * , x 0 * ) for later times.

Theorem 2.3. Suppose

(i) x 0 is a regular singular point and

(ii) ( x 0 − x ) p ( x ) and ( x 0 − x ) 2 q ( x ) are analytic at x 0

Then Frobenius method is effective at regular singular point of the form.

( x − x 0 ) ∑ n = 0 ∞ a n ( x − x 0 ) n

Theorem 2.4. Assume that x 0 = 0 otherwise the point x = x 0 is a singular point of

P ( x ) y ″ ( x ) + Q ( x ) y ′ + R ( x ) y ( x ) = 0 (5)

and that Q 1 ( x ) = x Q ( x ) P ( x ) and Q 2 ( x ) = x 2 R ( X ) P ( x ) are analytic at x = 0 , then they will have Maclaurin series expansion

f ( x ) = f ( 0 ) x + f ″ ( 0 ) 2 ! + f 3 ( 0 ) 3 ! x 3 + ⋯ + f n ( 0 ) n ! x n + ⋯ (6)

with radius of convergence r 1 > 0 and r 2 > 0 respectively. That is Q 1 ( x ) = x Q ( x ) P ( x ) = ∑ n = 0 ∞ P n X n which converges for | X | < r 2 . Then the point x 0 = 0 is called a regular singular point of (5).

Theorem 2.5. Consider a power series

f ( x ) = ∑ n = 0 ∞ a n x n (7)

with radius of convergence R, then term by term differential and integration of the power series is permitted and does not change the radius of convergence that is;

d f d x = d d x ∑ n = 0 ∞ a n x 0 = ∑ n = 0 ∞ d d x a n x n = ∑ n = 1 ∞ n a n x n − 1 , | x | < R (8)

∫ f ( x ) d x = ∫ d x ∑ n = 0 ∞ a n x n = ∑ n = 0 ∞ ∫ d x a n x n = ∑ n = 0 ∞ a n x n + 1 n + 1 , | x | < R (9)

Theorem 2.6. Let A be an n ∗ n matrix and let the eigenvalue of A be denoted by λ ( A ) and consider the linear system of differential equation

x ˙ ( t ) = A x ( t ) , x ∈ R n (10)

(i) If R e ( λ ( A ) ) ≤ 0 and all the eigenvalues of A with real part zero are simple, then zero is a stable fixed point of (10)

(ii) If R e λ ( A ) < 0 , then zero is a globally asymptotically stable solution of (10)

(iii) If there is an eigenvalue of A with positive real part, then zero is unstable

We consider the equation

d 2 x d t 2 + p x = 0 (11)

where p is a positive constant. Equation (11) can be written as

x ¨ + p x = 0 (12)

Let x 1 = x , x ˙ 1 = x 2 , x ¨ 1 = x ˙ 2

then equation (12) can be written as

x ˙ 2 + p x 1 = 0 (13)

The equivalent system is given by

x ˙ 1 = x 2 x ˙ 2 = − p x 1 (14)

(14) can be reduced in matrix form as

( x ˙ 1 x ˙ 2 ) = ( 0 1 − p 0 ) ( x 1 x 2 ) (15)

(15) can further be written as

x ˙ = A x

where, x ˙ = ( x ˙ 1 x ˙ 2 ) , x = ( x 1 x 2 ) and A = ( 0 1 − p 0 )

For the eigenvalue of A we compute

| A − I λ | = 0 (16)

| A − I λ | ⇒ | ( 0 1 − p 0 ) − λ ( 1 0 0 1 ) | = 0 (17)

⇒ | ( 0 1 − p 0 ) − ( λ 0 0 λ ) | = 0

⇒ | − λ 1 − p − λ | = 0 (18)

λ 2 + p = 0 (19)

Evaluating λ in (19) we have

λ = ± i p 1 2 (20)

Since the eigenvalue is in complex form with real part equals to zero, then the equilibrium point is unstable.

We consider the Mathieu equation of the form;

d 2 x d t 2 + p x = 0 (21)

From (21) we assume a power series solution of the form;

x = ∑ n = 0 ∞ a n t n + k (22)

Differentiating (21) term by term we have

x ˙ = ∑ n = 0 ∞ ( n + k ) a n t n + k − 1 (23)

x ¨ = ∑ n = 0 ∞ ( n + k ) ( n + k − 1 ) a n t n + k − 2 (24)

Substituting (23) and (24) into (21) we have

∑ n = 0 ∞ ( n + k ) ( n + k − 1 ) a n t n + k − 2 + p ∑ n = 0 ∞ a n t n + k = 0 (25)

The indicial equation is obtained by looking at the coefficient of the lowest power x k − 2 . Since this only occur in the first sum (for n = 0), it must vanish. This reduces (25) to

k ( k − 1 ) a 0 t k − 2 = 0 (26)

There are therefore two possible values of the index k = 0, k = 1 and this is quite typical for a second order equation. Changing the dummy index in the first sum by n → n + 2 we have

∑ n = − 2 ∞ ( n + k + 2 ) ( n + k + 1 ) a n + 2 t n + k + p ∑ n = 0 ∞ a n t n + k = 0 (27)

All the powers now look alike and we can now compare coefficient to obtain the recurrence relation given by

( n + k + 2 ) ( n + k + 1 ) a n + 2 + p a n = 0 (28)

which reduces to

a n + 2 = − p ( n + k + 2 ) ( n + k + 1 ) a n (29)

Given the value of a 0 , we can evaluate a 2 , a 4 , etc. The odd a n are completely independent and as far as getting a solution is concerned, we can put them all to zero. This independent of the odd and even a n is a consequence of the fact that odd and even solution of the differential equation are possible.

In order to generate these odd/even solution, it is easiest to put a 1 = 0 in order not to create extra solution by merely using some of the key solution into that of k = 0 .

The recurrence relation for k = 0 given by

a n + 2 = − p ( n + 2 ) ( n + 1 ) a n (30)

has the solution

x ( t ) = a 0 cos t + a 1 sin t (31)

The recurrence relation for k = 1 is

a n + 2 = − p ( n + 3 ) ( n + 2 ) a n (32)

with the solution

x ( t ) = a 1 sin t + a 0 cos t (33)

Recall that

cos t = ∑ n = 0 ∞ ( − 1 ) t 2 n ( 2 n ) ! = 1 − t 2 2 ! + t 4 4 ! − ⋯ (34)

sin t = ∑ n = 0 ∞ ( − 1 ) n t 2 n + 1 ( 2 n + 1 ) ! = t − t 3 3 ! + t 5 5 ! − ⋯ (35)

Combining equations (31) and (33) we have

x ( t ) = a 0 cos t + a 1 sin t

where a 0 and a 1 are arbitrary parameter

Using Ratio Test on cos t , we have

lim n → ∞ | a n + 1 a n | = lim n → ∞ | t 2 n + 2 ( 2 n + 2 ) ! ⋅ ( 2 n ) ! t 2 n | (36)

= lim n → ∞ | t 2 ( 2 n + 2 ) ! ⋅ ( 2 n ) ! ( 2 n + 1 ) ( 2 n ) ! | = lim n → ∞ 1 4 n 2 − 6 n + 2 | t 2 |

= lim n → ∞ 1 4 n 2 − 6 n + 2 | t 2 | = 0 (37)

Similarly for sin x = ∑ n = 0 ∞ ( − 1 ) n t 2 n + 1 2 n + 1 we have

lim n → ∞ | a n + 1 a n | = lim n → ∞ | t 2 n + 2 ( 2 n + 3 ) ! ⋅ ( 2 n + 1 ) ! t 2 n − 1 | = lim n → ∞ | t 2 ( 2 n + 3 ) ( 2 n + 2 ) | = 0 (38)

Thus the interval of convergence is ( ∞ , − ∞ ) , radius = infinity

Therefore

x ( t ) = a 0 cos t + a 1 sin t = a 0 ⋅ 0 + a 1 ⋅ 0 = 0 (39)

Since the limit of the odd and even function is equal to zero we conclude that the solution of Mathieu equation converges. Applying the boundary condition we have that

x ( 0 ) = x ( 2 π ) = a 1 (40)

p = 0.01

Define a function that determines a vector of derivatives values at any solution point ( t , X )

D ( t , X ) : = [ X 1 − 100 X 0 − p ( X 0 ) 2 ]

Define an additional argument for solving the ODE

t 0 : = 0 Initial value of independent variable

t 1 : = 10 Final value of independent variable

X 0 : = ( 0 1 ) Vector of initial function values

N : = 1500 Numbers of solution values on [ t 0 , t 1 ]

S : = ( X 0 , t 0 , t 1 , N , D )

t : = S 〈 0 〉 Independent variables values

X 1 : = S 〈 1 〉 First solution function values

X 2 : = S 〈 2 〉 Second solution function values

From our result, we observed that the solution existed using the Frobenius method and also periodic. The solution converged at the equilibrium point but unfortunately this convergence did not imply asymptotic stability, and the converse is true. The solution was observed to be unbounded for the given parameters w-ε. Since the solution is unbounded, we concluded that the corresponding equilibrium point in w-ε plane is unstable. This can be seen in

In _{0} and the constant p. The periodic nature shown is unstable since the trajectory did not start from the equilibrium point. The maximum displacement reached by the trajectory is at ±0.1 representing a value along the first solution function values.

0 | 1 | 2 | |
---|---|---|---|

0 | 0 | 0 | 1 |

1 | 6.667 | 6.662 | 0.998 |

2 | 0.013 | 0.013 | 0.991 |

3 | 0.02 | 0.02 | 0.98 |

4 | 0.027 | 0.026 | 0.965 |

5 | 0.033 | 0.033 | 0.945 |

6 | 0.04 | 0.039 | 0.921 |

7 | 0.047 | 0.045 | 0.893 |

8 | 0.053 | 0.051 | 0.861 |

9 | 0.06 | 0.056 | 0.825 |

10 | 0.067 | 0.062 | 0.786 |

11 | 0.073 | 0.067 | 0.743 |

12 | 0.08 | 0.072 | 0.697 |

13 | 0.087 | 0.076 | 0.647 |

14 | 0.093 | 0.08 | 0.595 |

15 | 0.1 | 0.084 | ... |

In

In _{2} and X_{1}. The phase portrait was seen to be far away from the equilibrium point. This shows that the solution of Mathieu’s equation is unstable with the maximum displacement coinciding with the maximum displacement of X_{1} in

The authors declare no conflicts of interest regarding the publication of this paper.

Eze, E.O., Obasi, U.E., Ujumadu, R.N. and Kalu, G.I. (2020) Maximum Interval of Stability and Convergence of Solution of a Forced Mathieu’s Equation. World Journal of Mechanics, 10, 210-219. https://doi.org/10.4236/wjm.2020.1011015