Problem
Consider the differential equations
dx
1 3 2 2 3
——— = - x + a x + a x x + a x x + a x
dt 2 0 1 1 1 2 2 1 2 3 2
dx
2 3 2 2 3
——— = x + b x + b x x + b x x + b x
dt 1 0 2 1 2 1 2 2 1 3 1
Show that the solution
x
= x = 0 is asymptotically stable (i.e., there exists a δ > 0 such that for every solution x = x(t) with | x(0) | < δ we have x(t) = o )
if
3a
+ 3b
+ a
+ b < 0
and completely unstable (i.e., there exists a δ > 0 such that for every solution x = x(t) with x(0) ≠ o there exists a T ≥ 0 such that | x(t) | ≥ δ for t ≥ T)
if
3a
+ 3b
+ a
+ b > 0.
G.W. Veltkamp.
Solution
Consider the function
2 2 1 4 1 4 3 3 1 3 3
V(x) = x + x + -(a + b )x - -(a + b )x - (a - b )(x x + x x ) + -(a + b )(x x - x x ).
- 1 2 2 1 3 1 2 3 1 2 0 0 1 2 1 2 3 2 2 1 2 1 2
Then along a trajectory
∂V dx ∂V dx
. d 1 2 1 4 4
V(t) = —— V(x(t)) = ——— ——— + ——— ——— = [a + b + —(a + b )](x +x ) + terms of degree 6.
dt - ∂x dt ∂x dt 0 0 3 2 2 1 2
1 2
It is obvious that V is a Lyapunov-function, i.e., V is continuously differentiable, V(o) = 0, V(x) > 0 in an open neighbourhood Ω(0 < |x| < δ) of x = o and (t) ≶ 0 for x(t) ∈ Ω if
a
+ b
+ (a
+ b) ≶ 0.
The statements concerning stability and instability now follow from well-known theorems of Lyapunov (cf., e.g., La Salle and Lefschetz, Stability by Lyapunov's Direct Method, N.Y., 1961).
Note. | The basic idea of the proof of Lyapunov's theorem is the following.
Let 0 <
δ <
δ , with sufficiently small
δ. Then the inequalities
δ ≤ V(x) ≤
δ
define a closed annular subdomain Ω of Ω which contains the origin inside its inner boundary.
If < 0 for x(t) ∈ Ω then ≤ -ε < 0 as long as x(t) ∈ Ω .
Hence if x(0) ∈ Ω then certainly V(x(t)) <
δ
for t >
(a-δ)/ε .
Hence
V(x(a)) = 0, which implies
x(t) = o. And similarly in the case of instability.
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