control theorydynamical systemsordinary differential equations
Consider the planar ODE
$\dot x_1 = x_2$
$\dot x_2 = – x_1^2 – 2 x_1 – 1$
Obliviously, $(x_1,x_2)=(-1,0)$ is an equilibrium point. The Jacobian matrix at this point is
$$J = \begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}$$
Thus, linearizarion fails in determining the stability. How can we determine the stability of this equilibrium point?
To find the equilibrium points, we want to find the points where $x'_1$ and $x'_2$ are zero.
So, we want to solve:
$\tag 1 x_1-x_2-x_1^3 = 0$
$\tag 2 x_1+x_2-x_2^3=0$
Using $(2)$, we have $x_1 = x_2^3 - x_2$ and substituting this into equation $(1)$ yields:
$$x_2^3 - x_2 - x^2 -(x_2^3 - x_2)^3 = -x_2 (x_2^8-3 x_2^6+3 x_2^4-2 x_2^2+2) = 0$$
If we substitute these back into $x_1 = x_2^3 - x_2$, we get the corresponding point as $0$. So, our critical point is $(0,0)$.
Now, you can use these two points with the Jacobian (assuming they are well behaved, do you know when you can use linearization and when you cannot) and determine how these equilibria behave.
Hint: Look at these two points in the phase portrait and your analysis should show this.
Hint.
A good stream plot can help a lot!
In red the equilibrium points for $\epsilon = 0.1$.
The corresponding MATHEMATICA script.
epsilon = 0.1;
str = StreamPlot[{x (x - 0.5) (x + 0.5) + epsilon y,
y (y - 0.5) (y + 0.5) + epsilon x}, {x, -1, 1}, {y, -1, 1},
MeshFunctions -> Function[{x, y, vx, vy, n}, n], Mesh -> 55];
sols = Quiet[Solve[{x (x - 0.5) (x + 0.5) + epsilon y == 0,
y (y - 0.5) (y + 0.5) + epsilon x == 0}, {x, y}]];
equil = Table[Graphics[{Red, Disk[({x, y} /. sols[[k]]), 0.02]}], {k, 1,
Length[sols]}];
Show[str, equil]
Best Answer
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.