# Showing that all solutions to the ODE $x''=4x^3$ cease to exist after a finite time

I am given the "Newton equation" $$x''(t)=4x^3(t)$$ The first part asks for finding a "conservation of energy" law. I'll spare the details:
the conclusion is that every solution has a constant $C$ such that $\frac{y^2}{2}-x^4=C$ where $y(t)=x'(t)$. (That follows from writing the ODE as a system of two ODEs: $x'=y$ and $y'=4x^3$ which is exact).

The second part asks to show that except the trivial solution $x\equiv0$, all solutions cease to exist (and therefore tend to infinity) after a finite time.
Now, this is something that can't be inferred by knowing the relation between $x$ and $y$ because it relies on their dependence on $t$, so I suspect I don't have the means to solve it.

I'd like to hear what is the technique for these questions so I can do the rest of the problems in my homework that contain similar questions (like, which solutions are bounded).

Thanks!

## 2 Answers

For fixed C, once x is large enough y is bounded from below by $x^2$, so the time it takes for $x$ to go from $n$ to $n+1$ is at most $n^{-2}$. This is summable over $n$, therefore there is a time $T$ such that $x\to\infty$ as $t\to T^-$. The same if they are negative.

Now you have to show that the solution eventually leaves a finite region so you can apply this observation which requires x large enough. If they go in the negative direction a similar argument works.

If $C\ne 0$ you can show that $\|(x',y')\|$ is bounded from below on the entire curve so it has to go away.

Also, notice that not all solutions cease to exist after a finite time. Taking $C=0$, you have $y=2x^2$ for $x<0$ and $y=-2x^2$ for $x>0$ which approximate the fixed point at $t\to\infty$. More explicitly, $x=\pm \sqrt 2 (t-t_0)^{-1}$ are solutions that are defined for all positive times.

Since $$y=\frac{dx}{dt}\\ t=\int_{x(0)}^\infty \frac{dx}y\\ =\int_{x(0)}^\infty \frac{dx}{\sqrt{2C+2x^4}}$$ and you need to show this is finite.

• Oh I see, but why did you choose 0 and infinity as limits? And isn't the integral divergent for $c\leq 0$? – 35T41 Dec 9 '17 at 11:48
• I guess the lower limit should depend on C, it is really $x(0)$. You want $x$ to increase to $\infty$, in a finite time, so $x=\infty$ is the other limit. – Michael Dec 9 '17 at 13:22