Nature of critical point $(0,0)$ for the given 2nd order ODE Consider the 2nd order ODE: $$\frac{d^2y}{d^2t}+2\alpha \frac{dy}{dt}+\beta^2 y=0$$ where $\alpha >\beta>0$.
Then find the nature of the critical point of the above ODE.
First I put $\frac{dy}{dt}=x$ and $\frac{dx}{dt}=-2\alpha x-\beta^2y$. Then $(0,0)$ is a critical point and the Jacobian matrix is $$J=\left(\begin{matrix} 1 & 0\\-2\alpha & -\beta^2\end{matrix}\right) $$
Roots of the characteristic equation of $J$ are $1$ and $-\beta^2$, which are real and opposite sign. So, the critical point $(0,0)$ is a saddle point.
Is it correct ?
 A: We are given the ODE
$$\frac{d^2y}{d^2t}+2\alpha \frac{dy}{dt}+\beta^2 y=0, ~~~\text{with}~~~\alpha >\beta>0$$
We can write this as a system of first order ODEs by letting $x_1 = y$
$$\begin{align}x_1' &= y' = x_2 \\ x_2' &= y'' = -\beta^2 x_1 - 2 \alpha x_2\end{align}$$
As a system we have
$$X'(t) = \begin{bmatrix} x_1'(t) \\ x_2'(t) \end{bmatrix} = \begin{bmatrix} 0 & 1 \\- \beta^2 & -2 \alpha \end{bmatrix}\begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix}$$
The critical points are found by simultaneously solving for $x_1' = x_2' = 0$, which give
$$(x_1, x_2) = (0, 0)$$
Since this is a linear system, the Jacobian matrix at that critical point is
$$J(x_1, x_2) = J(0, 0) =  \begin{bmatrix} 0 & 1 \\- \beta^2 & -2 \alpha  \end{bmatrix}$$
The eigenvalues are given by
$$\lambda_1 = - \alpha -\sqrt{ \alpha ^2-\beta^2}, \lambda_2 =- \alpha + \sqrt{ \alpha ^2-\beta^2}$$
We also have the constraint that $a \gt \beta > 0$.
Can you continue?
A: For linear systems, there are a few tools that can be used. The other two approaches do not require the computation of the eigenvalues of the system. The ODE
$$ \ddot{y} + 2 \alpha \dot{y} + \beta^{2} y = 0 $$
can be normalized by $\beta^{2}$ to become
$$ \frac{1}{\beta^{2}} \ddot{y} + 2 \frac{\alpha}{\beta^{2}} \dot{y} + y = 0. $$
Taking the Laplace transform, term by term, and you get
$$ \frac{s^{2}}{\beta^{2}} + 2 \frac{\alpha}{\beta} \frac{s}{\beta} + 1 = 0, $$
which can be rewritten in this form
$$ \bar{s}^{2} + 2 \frac{\alpha}{\beta} \bar{s} + 1 = 0, $$
where $\bar{s} = \frac{s}{\beta}$. In this form, if the Routh–Hurwitz stability criterion is applied, then it can be easily shown that the stability criterion is
$$ 2 \frac{\alpha}{\beta} > 0 $$
Since it is known that $\alpha > \beta > 0$, therefore the criterion is satisfied and the critical point of the system is a stable equilibrium point. In fact, this is consistent with the example for the second-order linear system given here.

Another approach is to apply the Lyapunov stability theorem, which generally says that a system with the origin $\mathbf{y} = \mathbf{0}$ is asymptotically stable if the following conditions are satisfied:

*

*$V\left(\mathbf{0}\right) = 0,$

*$V\left(\mathbf{y}\right) > 0, \quad \forall \; \mathbf{y} \neq \mathbf{0},$

*$\dot{V}\left(\mathbf{y}\right) < 0, \quad \forall \; \mathbf{y} \neq \mathbf{0}.$
Consider the quadratic Lyapunov function candidate:
$$ V\left(y, \dot{y}\right) = \frac{1}{2} \left[\matrix{y & \dot{y}}\right] \left[\matrix{4 \alpha^{2} + 2 \beta^{2} & 2 \alpha \cr 2 \alpha & 2}\right] \left[\matrix{y \cr \dot{y}}\right], $$
$$ V\left(y, \dot{y}\right) = \frac{1}{2} \left(4 \alpha^{2} + 2 \beta^{2}\right) y^{2} + 2 \alpha y \dot{y} + \dot{y}^{2}. $$
The first condition $V(\mathbf{0}) = 0$ is satisfied when $\left(y, \dot{y}\right) = \left(0, 0\right)$. The second condition $V(\mathbf{y}) > 0$ is also satisfied when $\alpha > 0$ and $\beta > 0$.
Taking the time derivative for $V(\mathbf{y})$
$$ \dot{V}\left(y, \dot{y}\right) = 4 \alpha^{2} y \dot{y} + 2 \beta^{2} y \dot{y} + 2 \alpha y \ddot{y} + 2 \alpha \dot{y}^{2} + 2 \dot{y} \ddot{y}, $$
$$ \dot{V}\left(y, \dot{y}\right) = 4 \alpha^{2} y \dot{y} + 2 \beta^{2} y \dot{y} + 2 \alpha y \left(- \beta^{2} y - 2 \alpha \dot{y}\right) + 2 \alpha \dot{y}^{2} + 2 \dot{y} \left(- \beta^{2} y - 2 \alpha \dot{y}\right), $$
$$ \dot{V}\left(y, \dot{y}\right) = 4 \alpha^{2} y \dot{y} + 2 \beta^{2} y \dot{y} - 2 \alpha \beta^{2} y^{2} - 4 \alpha^{2} y \dot{y} + 2 \alpha \dot{y}^{2} - 2 \beta^{2} y \dot{y} - 4 \alpha \dot{y}^{2}. $$
Cancelling out common factors
$$ \dot{V}\left(y, \dot{y}\right) = - 2 \alpha \beta^{2} y^{2} - 2 \alpha \dot{y}^{2}. $$
Since $\alpha > \beta > 0$, therefore the third condition $V(\mathbf{y}) < 0$ is satisfied and the critical point of the system is asymptotically stable.
