Investigate the stability of the equilibrium point (0,0) by constructing a suitable Lyapunov function. I'm struggling to follow along with any online notes, as I understand there is no general method for constructing a Lyapunov function, any help would be appreciated.
Consider the following dynamical system
$\dot x = −x + 4y$ , $\dot y = −x − y^3$
Investigate the stability of the equilibrium point $(0,0)$ by constructing a suitable Lyapunov function. [Hint: try a function of the form $L(x, y) = x^2 + ay^2$]
 A: With
$\dot x = -x + 4y \tag 1$
and
$\dot y = -x - y^3 \tag 2$
and
$L(x, y) = x^2 + ay^2, \tag 3$
we have
$\dot L(x, y) = \dfrac{L(x, y)}{\partial x} \dot x + \dfrac{L(x, y)}{\partial y} \dot y$
$= (2x)(-x + 4y) + (2ay)(-x - y^3)$
$= -2x^2 + 8xy - 2axy - 2ay^4$
$= -2(x^2 + ay^4) + (8 - 2a)xy; \tag 4$
if we choose
$a = 4, \tag 5$
then
$8 - 2a = 0, \tag 6$
and hence (4) becomes
$\dot L(x, y) = -2(x^2 + 4y^4) < 0 \tag 7$
for all
$(x, y) \ne (0, 0); \tag 8$
furthermore
$L(0, 0) = 0, \tag 9$
and
$L(x, y) = x^2 + 4y^2 > 0 \tag{10}$
whenever (8) binds. Thus $L(x, y)$ as in (10) is a Lyapunov function for the system (1)-(2), and the origin $(0, 0)$ is a stable equilibrium point.
Note Added in Edit, Sunday 29 November 2020 9:46 AM PST: $(0, 0)$ is in fact the unique equilibrium point of (1)-(2), for with
$\dot x = \dot y = 0, \tag{11}$
equations (1), (2) yield
$-x + 4y = 0, \tag{12}$
$-x - y^3 = 0; \tag{13}$
from (12),
$x = 4y, \tag{14}$
and substituting this into (13) we find
$-4y - y^3 = 0, \tag{15}$
or
$y(y^2 + 4) = 0; \tag{16}$
for every real $y$,
$y^2 + 4 > 0, \tag{17}$
which forces
$y = 0; \tag{18}$
then both (12) and (13) affirm that
$x = 0 \tag{19}$
as well, and we see that $(0, 0)$ is the sole equilibrium point of (1)-(2).  End of Note.
