Show that if $ \lim _{|x| \rightarrow \infty} \int_0^x h(s) d s=+\infty $ then all solutions to this ODE are bounded.

Question

a) Analyze $x^{\prime \prime}+f(x) x^{\prime}+h(x)=0$ where $f(x)>0$ and $x h(x)>0$ for $x \neq 0$ and such that $f, h$ are continuous.

b) Additionally, show that if $$ \lim _{|x| \rightarrow \infty} \int_0^x h(s) d s=+\infty $$ then all solutions to this ODE are bounded.

My try:

I was analyzing by creating one system of two 1st order eqns by taking

$$x'=y$$ and $$y'=-fy-h.$$

Then I tried a lot of different ways but can't find a Lyapunov function so that it would be asymptotically stable. Then I am not having any clue about the second part because I am unable to get the Lyapunov function.

If I get the Lyapunov then I can set the level sets of the Liapunov function and that can enclose any region in the plane that includes the origin. Please help.

Answer 1

If you think of the system $x''+f(x)x'+h(x)=0$ as describing a physical nonlinear oscillator, then $-f(x)x'$ and $-h(x)$ would be the (nonlinear) dissipative and restoring forces respectively. $xh(x)>0$ for all nonzero $x$ is equivalent to saying that $$\begin{cases}h(x)>0 & x>0\\h(x)<0 &x<0\end{cases}$$ so that the "restoring force" $-h(x)$ will always point back towards the origin, and with the dissipative term one can intuitively expect all initial conditions to settle at the origin eventually. Also worth noting is that continuity implies $h(0)=0$ and that $h$ is integrable.

Lyapunov functions essentially generalise the notion of energy from physical dynamical systems, so trying to study the corresponding "total energy" function for this oscillator would be a good place to start. The potential energy from the restoring force is $\int_0^xh(x)\text{ d}x$ and the kinetic energy is proportional to $y^2$ (if we define $y=x'$). Therefore, we should consider $$V=\int_0^xh(x)\text{ d}x+ay^2$$ where a is a constant we can pick later. Differentiating everything and substituting the original system of equations $$\begin{cases}x'=y\\y'=-f(x)y-h(x)\end{cases}$$ gives $$V'=(1-2a)h(x)y-2af(x)y^2$$ Since $f(x)>0$, if we pick $a$ such that the first term vanishes (i.e. $a=1/2$), then $V=\int_0^xh(x)\text{ d}x+\frac12y^2$ and $$V'=-f(x)y^2\le0$$and is strictly negative when $y\neq0$. However, this is only enough narrow down the possible accumulation points of solutions to the line $y=0$. (Thanks to @Kwin van der Veen for pointing this out!) To show that the origin is the only asympotically stable point, we need to use LaSalle's invariance principle by showing that $V>0$ everywhere besides the origin. This can be done by noting that the inequalities on $h(x)$ imply that $\int_0^xh(x)\text{ d}x>0$ for both positive and negative $x$. Together, this shows that $V$ is a Lyapunov function and that the origin is locally asymptotically stable.

The condition in (b) that $\int_0^xh(x)\text{ d}x\to\infty$ as $x\to\infty$ implies that $V(x,y)\to\infty$ as $\sqrt{x^2+y^2}\to\infty$, which implies that the origin is globally asymptotically stable as well.