Layla Velazquez

2022-06-25

I am given a single second order differential equation:
$\stackrel{¨}{x}-{x}^{3}-2{x}^{2}\stackrel{˙}{x}+1=0$
.. 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:
$\stackrel{˙}{{x}_{1}}={x}_{2}$
$\stackrel{˙}{{x}_{2}}-{x}_{1}^{3}-2{x}_{1}^{2}{x}_{2}+1=0$
Thereafter, I follow the same process as starting off with a set of two first order differential equations. That is, set ${\stackrel{˙}{x}}_{1}$ and ${\stackrel{˙}{x}}_{2}$ 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 & Explanation

Dustin Durham

The differential system is:
$\left\{\begin{array}{l}{\stackrel{˙}{x}}_{1}={x}_{2}\\ \stackrel{˙}{{x}_{2}}={x}_{1}^{3}+2{x}_{1}^{2}{x}_{2}-1\end{array}$
You have a critical point at $\left({x}_{1},{x}_{2}\right)=\left(1,0\right)$. Change the variables:
$\left\{\begin{array}{l}{x}_{1}={y}_{1}+1\\ {x}_{2}={y}_{2}\end{array}$
The jacobian matrix is:
$J=\left(\begin{array}{cc}0& 1\\ 3\left({y}_{1}+1{\right)}^{2}+4\left({y}_{1}+1\right){y}_{2}& 2\left({y}_{1}+1{\right)}^{2}\end{array}\right)$
Now the critical point is at $\left({y}_{1},{y}_{2}\right)=\left(0,0\right)$
$J=\left(\begin{array}{cc}0& 1\\ 3& 2\end{array}\right)$
The linearized system now is:
${Y}^{\prime }=\left(\begin{array}{cc}0& 1\\ 3& 2\end{array}\right)Y$
You compute the eigenvalues:
$P\left(\lambda \right)=|\begin{array}{cc}-\lambda & 1\\ 3& 2-\lambda \end{array}|$
$P\left(\lambda \right)={\lambda }^{2}-2\lambda -3=0$
${\lambda }_{1,2}=-1,3$
So the critical point is a saddle point.

Do you have a similar question?

Recalculate according to your conditions!