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
 1
= x
 2
= 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
 lim t→∞
x(t) = o )

if              3a
 0
+ 3b
 0
+ a
 2
+ b
 2
< 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   tT)

if              3a
 0
+ 3b
 0
+ a
 2
+ b
 2
> 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
 . V
(t) 0
for   x(t) Ω   if   a
 0
+ b
 0
+
 13
(a
 2
+ b
 2
) 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
 1
< δ
 2
,
with sufficiently small   δ
 2
. Then the inequalities
δ
 1
V(x) ≤ δ
 2

define a closed annular subdomain   Ω
 1
of   Ω   which contains the origin inside its inner boundary.
If
 . V
< 0
for   x(t) Ω   then
 . V
≤ -ε < 0
as long as   x(t) Ω
 1
.

Hence if   x(0) Ω
 1
then certainly   V(x(t)) < δ
 1
for   t > (a
 2
-δ
 1
)/ε .

Hence
 lim t→∞
V(x(a)) = 0,
which implies
 lim t→∞
x(t) = o.
And similarly in the case of instability.