# Classification of Critical Points of Second Order Differential Equation

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?

• You could also use $v=\dot x-\frac23x^3$ which then gives $\dot v=x^3-1$ for a "more even" distribution of the terms of the equation. May 25, 2020 at 10:59

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.
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.
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.
• I feel embarrassed to ask this, but would it be possible to edit your answer to explain how we get $x=1+u$ instead of $x=1$? I also don't understand how the equation following this is derived. May 25, 2020 at 12:30
• You want to explore how solutions close to the stationary one behave, so if $x$ is close to $1$ then $x=1+u$ with $u=x-1\approx 0$, $\dot u\approx 0$ and $|u|^2\ll|u|$ etc. The constant terms cancel, and the quadratic and higher degree terms are very small against the remaining linear terms. So in a first approximation, retain only the terms linear in $u$. May 25, 2020 at 12:58