The critical point $(0,0)$ is assymptotically stable. Demostrate $a_{11}+a_{22}<0$ and $a_{11}a_{22}-a_{12}a_{21}>0.$

Question

I've got the following linear system:

$$\frac{dx}{dt}=a_{11}x+a_{12}y$$ $$\frac{dy}{dt}=a_{21}x+a_{22}y$$

The critical point $(0,0)$ is an assymptotically stable critical point of the system.

We have to demostrate that $$a_{11}+a_{22}<0$$ and $$a_{11}a_{22}-a_{12}a_{21}>0.$$

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I have writted the system like this:

$$\begin{pmatrix} x' \\ y' \end{pmatrix}=\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$$

If we call $A=\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{pmatrix}$

$Traza(A)=a_{11}+a_{22}$

and

$Det(A)=a_{11}a_{22}-a_{12}a_{21}$

So now, I have to do something with the eigenvalues, true?

How can I follow? I'm a bit lost.

Answer 1

I suppose that the given system has real coefficients.

That is the trick I remember: $\dot x=a_{11}x+a_{12}y\implies\ddot x=a_{11}\dot x+a_{12}\dot y\implies\ddot x=a_{11}\dot x+a_{12}(a_{21}x+a_{22}y)\implies\\\ddot x=a_{11}\dot x+a_{12}a_{21}x+a_{22}a_{12}y\implies\ddot x=a_{11}\dot x+a_{12}a_{21}x+a_{22}(\dot x-a_{11}x)\implies\\\ddot x=(a_{11}+a_{22})\dot x+(a_{12}a_{21}-a_{11}a_{22})x$

Hence we get a $2$nd order linear differential equation $$\ddot x-(a_{11}+a_{22})\dot x+(a_{11}a_{22}-a_{12}a_{21})x=0\tag{LDE}$$ with characteristic equation $m^2-Tr(A)m+\Delta(A)=0$.

We have two cases:

1. There are two real roots, $\lambda_1,\lambda_2$ and the solution of the LDE is $x(t)=c_1e^{\lambda_1t}+c_2e^{\lambda_2t}$. For asympotic stability (AS), $x(t)\rightarrow 0$ when $t\rightarrow\infty$. Assuming the general case of the solution, this happens iff $\lambda_1,\lambda_2<0$ iff $Tr(A)=\lambda_1+\lambda_2<0$ and $\Delta(A)=\lambda_1\lambda_2>0$.
2. There are two complex-conjugate roots $\alpha\pm\beta i$ and the solution is $x(t)=e^{\alpha}(c_1\cos\beta t+c_2\sin\beta t)$. Again, the solution is AS iff $\alpha=Tr(A)/2<0\implies Tr(A)<0.$ In this case, $\Delta(A)=\alpha^2+\beta^2>0.$

Answer 2

Let's take an alternative approach without solving for the eigenvalues. The phrase "asymptotically stable" implies that Lyapunov's method can be used, in which the following conditions must be satisfied:

1. $ V\left(\mathbf{0}\right) = 0 $
2. $ V\left(\mathbf{x}\right) > 0, \quad \forall \; \mathbf{x} \neq \mathbf{0} $
3. $ \dot{V}\left(\mathbf{x}\right) < 0, \quad \forall \; \mathbf{x} \neq \mathbf{0} $

Consider the Lyapunov function candidate:

$$ V\left(\mathbf{x}\right) = - \frac{1}{2} a_{21} x^{2} + \frac{1}{2} a_{12} y^{2}. $$

The first condition is satisfied when $(x, y) = (0, 0)$. In order to satisfy the second condition $V\left(\mathbf{x}\right) > 0$, then we must have

$$ a_{12} > 0, $$

$$ a_{21} < 0. $$

Taking the time derivative for $V\left(\mathbf{x}\right)$, we have

$$ \dot{V}\left(\mathbf{x}\right) = - a_{21} x \dot{x} + a_{12} y \dot{y} $$

$$ \dot{V}\left(\mathbf{x}\right) = - a_{21} x \left(a_{11} x + a_{12} y\right) + a_{12} y \left(a_{21} x + a_{22} y\right) $$

$$ \dot{V}\left(\mathbf{x}\right) = - a_{11} a_{21} x^{2} - a_{12} a_{21} x y + a_{12} a_{21} x y + a_{12} a_{22} y^{2}. $$

Cancelling like terms yields

$$ \dot{V}\left(\mathbf{x}\right) = - a_{11} a_{21} x^{2} + a_{12} a_{22} y^{2}. $$

In order to satisfy the third condition $\dot{V}\left(\mathbf{x}\right) < 0$, then we must have

$$ a_{11} a_{21} > 0, $$

$$ a_{12} a_{22} < 0. $$

From $a_{12} > 0$ and $a_{21} < 0$, it must be true that

$$ a_{11} < 0, $$

$$ a_{22} < 0. $$

Since $a_{11} < 0$ and $a_{22} < 0$, then

$$ a_{11} + a_{22} < 0. $$

Similarly, since $a_{11} a_{22} > 0$ and $a_{12} a_{21} < 0$, then

$$ a_{11} a_{22} - a_{12} a_{21} > 0. $$