Show that origin is globally asymptotically stable.
$$\begin{eqnarray} x' &=& −(x + y)^3\\ y' &=& (x − y)^3 \end{eqnarray}$$
I know to prove that $V'(x)$ has to be negative which I can prove. However, I can't seem to figure out how to get $V(x)$. Can anyone put me in right direction to how to calculate $V(x)$ for it.
On
As far as I am aware there are no general steps for finding Lyapunov functions, but for low dimensional systems one can look at the stream plot and try to find a contour line such that all stream lines cross that contour pointing inwards. This contour is defined by setting $V(x,y)$ equal to some constant. Often the Lyapunov function $V(x,y)=x^2+y^2$ can be a good starting guess, which would yield a circle as contour. If this doesn't work one could increase one or more of the powers from two to four, six, ect. which would make the contour more squarish and/or perform a linear coordinate transformation which can rotate and squash the contour. These alteration from the starting guess are usually enough when the time derivatives are polynomials, as is the case in your example.
On
Normally the characterization of a Lyapunov function is mostly an artisanal work. After a $\frac{\pi}{4}$ rotation introduced by the coordinates change
$$ u = x + y\\ v = x - y $$
we arrive at the equivalent system
$$ \dot u = v^3-u^3\\ \dot v = -(v^3+u^3) $$
Attached the flow chart for this system in blue, with a potential Lyapunov function showing in red the level curves.
As @Cesareo wrote, after the change of variables $$ u = x + y\\ v = x - y $$ the system has the form $$ \left\{\begin{array}{lll} \dot u &=& v^3-u^3\\ \dot v &=& -(v^3+u^3). \end{array}\right. $$ I can be seen that the Lyapunov function is $V(u,v)=u^4+v^4$: its derivative $$ \dot V= 4u^3\dot u+4v^3\dot v =4u^3(v^3-u^3)-4v^3(v^3+u^3) $$ $$ =-4u^6-4v^6$$ is negative definite.
Hence, $V(x,y)=(x+y)^4+(x-y)^4$.