Investigate the stability of the equilibrium point (0,0) by constructing a suitable Lyapunov function.

Question

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 $L(x, y) = x^2 + ay^2$]

Answer 1

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.