Let $f\in C^1(\mathbb{R} ,\mathbb{R} )$ and satisfying the condition: $$ \forall x >0, \quad f(x)>0, \forall x<0 , \quad f(x)<0 $$

Let $(\alpha, \beta) \in \mathbb{R^2}$.

1) Show that the maximal solution of the differential equation: $$ y''+f(y')+y=0, \quad y(0)=\alpha, \quad y'(0)=\beta $$ is defined on $[0,+\infty)$

2) What can we say about the asymptotic behavior when the weather tends to $+ \infty$?

My attempt (I know that is not enough):

Physically I know that $y''+y=0$ is an ODE for harmonic oscillator and geometrically we can draw the solution as a phase portrait.

By Cauchy-Lipshitz theorem we know that the maximal solution is defined and unique.

  1. What's happen when we introduce $f(y')$ ?
  2. Can we understand the phenomenon with a drawing/Physically?
  3. Does anybody know how to obtain a good asymptotic expression ?

1 Answer 1


If you write the equation as $y''=-y-f(y')$ you can see that $-y$ is represents a force proportional and opposite direction to the displacement (a recuperation force) and $f'(y')$ represents friction, dependent on the velocity. From the physical point of whew we expect that the movement will be attenuated becas of the friction.

What can we do with mathematics? Multiply the equation by $y'$: $$ y''\,y'+y'\,y=\frac12(y'^2+y^2)'=-y'\,f(y')\le0. $$ since $y'\,f(y')\ge0$. This implies that $y'^2+y^2$ is decreasing and $y'^2+y^2\le\alpha^2+\beta^2$ on he interval of existence. From this follows that the solution is defined on $[0,\infty)$.

To get an asymptotic estimate you can use the inequality $$ y(t)'^2+y(t)^2\le\alpha^2+\beta^2-\int_0^ty'(s)\,f(y'(s))\,dt. $$ You will probably need some assumption on $f$.

  • $\begingroup$ It's very interesting(+1). Why I will probably(?) need some assumption on f ? Perhaps We can estimate the asymptotic behavior without a rigorous asymptotic estimate. Perhaps study the adhesion values ​​of the parametric curve $\{(y'(t),y(t)), t\geq 0 \}$ What do you think about it? $\endgroup$
    – user119228
    Feb 27, 2014 at 18:15
  • $\begingroup$ Without further assumptions on $f$ I do not see how to prove the most basic result: $\lim_{t\to\infty}y(t)=0$ (this does not mean that it is not possible). May be you can find a Lyapunov function for the equivalent system $u'=v$, $v'=-u-f(v)$. $\endgroup$ Feb 27, 2014 at 18:23
  • $\begingroup$ hum..So I will search and try. $\endgroup$
    – user119228
    Feb 27, 2014 at 18:31
  • $\begingroup$ Integrate between $0$ and $t$ the equation multiplied by $y'$. $\endgroup$ Feb 27, 2014 at 22:15

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