Classification of Critical Points of Second Order Differential Equation

Question

I am given a single second order differential equation:

$\ddot{x}-x^3 - 2x^2\dot{x} + 1 = 0$

.. and am asked to classify the critical points as stable, unstable or saddle points.

Finding the critical points is an easy task for first order differential equation(s), both single equation and system of equations. However, I have never done so for second order, and higher, equations.

I have an idea on how to solve it but not sure the approach is correct. Am I correct in saying I need to split the single differential equation into two differential equations and rename the relevant terms? Doing so on the above equation gives:

$\dot{x_1} = x_2$

$\dot{x_2}-x_1^3-2x_1^2x_2+1 = 0$

Thereafter, I follow the same process as starting off with a set of two first order differential equations. That is, set $\dot{x}_1$ and $\dot{x}_2$ to $0$, and solve for the intersection of $x_1$ and $x_2$ to find the fixed points. The nature of the fixed points would then be determined by calculating the Trace and Determinant of the Jacobian at the specific fixed points.

Is my thinking on the right track?

Answer 1

The differential system is : $$ \begin{cases} \dot x_1=x_2 \\ \dot{x_2}=x_1^3+2x_1^2x_2-1 \end{cases} $$ You have a critical point at $(x_1,x_2)=(1,0)$. Change the variables: $$ \begin{cases} x_1=y_1+1 \\ {x_2}=y_2 \end{cases} $$

The jacobian matrix is: $$J=\pmatrix {0 & 1 \\ 3(y_1+1)^2+4(y_1+1)y_2 & 2(y_1+1)^2}$$ Now the critical point is at $(y_1,y_2)=(0,0)$ $$J=\pmatrix {0 & 1 \\ 3 & 2}$$ The linearized system now is: $$Y'=\pmatrix {0 & 1 \\ 3 & 2}Y$$ You compute the eigenvalues: $$P(\lambda)=\begin {vmatrix} -\lambda & 1 \\ 3 & 2-\lambda \end{vmatrix}$$ $$P(\lambda)=\lambda^2-2\lambda-3=0$$ $$\lambda_{1,2}=-1,3$$ So the critical point is a saddle point.

Answer 2

At critical, stationary points you have a solution that does not move, does not accelerate. Now insert $\dot x=0$, $\ddot x=0$ into the original equation to find that $-x^3+1=0$ remains.

Now I'm solving this differently than the task demands, the principal results however transfer.

You are probably only interested in the real solutions, so to explore the behavior of solutions close to the constant solution $x=1$ parametrize them as $x=1+u$, $\dot x=\dot u$, insert that into the equation, see that the constant terms cancel and ignore higher degree terms in $u$ retaining only the linear ones \begin{align} 0&=\ddot u-(1+3u\color{lightgray}{+3u^2+u^3}) -2(1\color{lightgray}{+2u+u^2})\dot u+1 \\&=\ddot u-2\dot u - 3u\color{lightgray}{+\text{higher degree terms}}. \end{align} The linearized equation now has a solution $u(t)=Ae^{3t}+Be^{-t}$ so that there are stable and unstable directions.