By Arthur Cayley

This quantity is made out of electronic photos from the Cornell collage Library ancient arithmetic Monographs assortment.

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12 [1] If a system has more than one critical point, then none of the critical points are globally asymptotically stable. 20 Show that the zero solution in the following systems is globally asymptotically stable. x1′ = − x2 − (1/2)x1x22 , x2′ = 3 x1 − (1/3)x2 . i. x1′ = x2 , x2′ = − x1 + a x12 x2 . a ≤ 0. ii. SOLUTION i. The origin is the only critical point of the system. Let us choose that V ( x) = a x12 + b x22 . Then, V *( x) = 2a x1 f1 + 2bx2 f 2 = 2a x1[− x2 − (1/2)x1x22 ] + 2bx2 [3 x1 − (1/3)x2 ] = 2x1x2 [− a + 3b] − ax12 x22 − (2/3)bx22 .

That is, every solution of the system irrespective of the initial value x0 converges to zero as t → ∞. We state the following theorem regarding the global stability of an equilibrium point. 7 [1] Assume that there exists a scalar function V(x) such that i. V is positive definite on Rn, and V(x) → ∞ as ǁ x ǁ → ∞, ii. V * (x) ≤ 0 on Rn. Then, all the solutions are bounded as t → ∞. This type of stability is also known as Lagrange stability. 13 is globally asymptotically stable. 13 is globally asymptotically stable.

Hence, construct a Lyapunov function, if possible, for the system. Is the origin (0, 0, 0) asymptotically stable? SOLUTION Set z = x1. Then, x1′ = x2 , x2′ = x3 , x3′ = −6 x3 − 11x2 − 6 x1 . We obtain the system of equations as x′ = Ax, where 0 A= 0 −6 1 0 −11 0 1 . −6 The characteristic equation of A is λ3 + 6λ2 + 11λ + 6 = 0. The eigenvalues are λ = −1, −2, −3. The matrix of eigenvectors (modal matrix) is obtained as 1 P = −1 1 1 −2 4 1 −1 −1 −3 and P AP = D = 0 0 9 The transformation x = Py gives y′ = Dy.