calculusordinary differential equationsstability-theory
I would like to determine the stability of equilibrium point $(x,0)$ of the differential equation
$\dot x = Ax$
$ A=
\bigg[
\begin{matrix}
0&0\\0&a
\end{matrix}
\bigg]
$ and $ a >0 $
I got
$ x'(t) = 0\ $
$\ y'(t) = a*y$
So, the equilibrium point is $(x,0)$.
How can I determine the stability of these equilibrium points?
Any help will be appreciated!
After the change of variables $y_1=x_1+1$, $y_2=x_2$ the system takes the form $$ \left\{\begin{array}{lll} \dot y_1&=&y_2\\ \dot y_2&=&-y_1^2.\\ \end{array}\right. $$ In order to prove the instability of the origin, one can use arguments similar to those used in the proof of the Chetaev instability theorem. Consider the set $$ G= \{ y\in\mathbb R^2:\; y_1<0,y_2<0 \}. $$ One can see that
But it means that no solution starting from the point $y^{start}$ in $G$ can stay in the $\|y^{start}\|$-neighborhood of the origin:
Moreover, let $y|_{t=0}=y^{start}=(y_1(0),y_2(0))$. We have $$ \forall t>0 \quad \dot y_1<y_2(0)<0,\; \dot y_2<-y^2_1(0)<0, $$ which implies $\lim_{t\to\infty} y_1(t)=-\infty$, $\lim_{t\to\infty} y_2(t)=-\infty$. Thus, the origin is unstable.
Best Answer
We want to determine the stability of the equilibrium points of the system $\dot x = Ax$, where
$$ A= \bigg[ \begin{matrix} 0&0\\0&a \end{matrix} \bigg], ~\text{with}~a >0 $$
The critical points are where we simultaneously have $x' = y' = 0$ and we get the entire $x-$axis as
$$(x, y) = (x, 0)$$
Since this system is decoupled, we can write $$\begin{align} x'(t) &= 0 \implies x(t) = c \\ y'(t) &= ay \implies y(t) = c e^{a t} \end{align}$$
The phase portrait is
Using all of the information above, we determine that the critical point is unstable.