Show that the zero solution of $$ \large \ddot{x}+bx^2 \dot{x}+kx=0$$ is asymptotically stable if $b>0$ and unstable if $b<0$. Does this depend on the sign of $k$?

I know for 1st order equation but how to find stability of 2nd order equation?

Can I convert it into system of 1st order equations as follows:

Let $y=x$ and $z=\dot x$. Then,

$\dot z=\ddot x=-bx^2 \dot x-kx=-by^2z-ky,$

i.e., $\dot z=-by^2z-ky, ......(1)$

and $ \dot y=z, .......(2)$

These are the two equations.

But how to find zero solutions and stability ?

Do we need to linearize this system again?

Help me


Any constant solution obviously has $\dot x=0$, $\ddot x=0$ so that the equation $kx=0$ remains.

For $k<0$ you get a saddle point, thus $k>0$.

For the stability consider $$ \frac{d}{dt}\frac12(\dot x^2+x^2)=-bx^2\dot x^2 $$ which tells you in which direction the solutions cross the circles around the origin.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.