MHB Determining Stability of Equilibrium Point for x'=y-x3-xy2, y'=-x-x2y-y3

onie mti
Messages
42
Reaction score
0
i have this system

x'=y-x3-xy2
y'=-x-x2y-y3

i worked it out and found the equilibrium point to be 0.

how do i determine whether it is stable, assymp stable or not stable
 
Physics news on Phys.org
onie mti said:
i have this system

x'=y-x3-xy2
y'=-x-x2y-y3

i worked it out and found the equilibrium point to be 0.

how do i determine whether it is stable, assymp stable or not stable

Let's define $x_{1}$ and $x_{2}$ the state variables, so that the system becomes...

$\displaystyle x_{1}^{\ '} = x_{2} - x_{1}^{3} - x_{1}\ x_{2}^{2}$

$\displaystyle x_{2}^{\ '} = - x_{1} - x_{1}^{2}\ x_{2} - x_{2}^{3}\ (1)$

You can choose as Lyapunov function $\displaystyle V(\overrightarrow x)= \frac{x_{1}^{2} + x_{2}^{2}}{2}$, so that is...

$\displaystyle V^{\ '} (\overrightarrow x) = x_{1}\ x_{2} - x_{1}^{4} - x_{1}^{2}\ x_{2}^{2} - x_{1}\ x_{2} - x_{1}^{2}\ x_{2}^{2} - x_{2}^{4} = - (x_{1}^{2} + x_{2}^{2})^{2} \le 0\ \forall \overrightarrow x\ (2)$

... so that the system is stable...

Kind regards

$\chi$ $\sigma$
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top