Oscillatory explosion

Nonlinear differential equations don’t necessarily have globally defined solutions, even if the equation is autonomous (no time is involved explicitly). The simplest example is ${y'=y^2}$ with initial condition ${y(0)=1}$ : the solution is ${y(t) = 1/(1-t )}$ , which ceases to exist at ${t=1}$ , escaping to infinity.

This kind of behavior can often be demonstrated without solving the ODE. Consider ${y''=y^3}$ with ${y(0)=1}$ , ${y'(0)=0}$ . I’ve no idea what the explicit solution is, but it’s clear that the quantity ${E(t) = \frac12 (y')^2 - \frac14 y^4}$ remains constant: indeed, ${E'(t) = y''y' - y^3 y' = 0}$ . (Here, ${E}$ is analogous to the sum of kinetic and potential energy in physics.)

The solution ${y}$ is increasing and convex. Let ${t_n}$ be such that ${y(t_n)=n}$ , e.g., ${t_1=0}$ . The invariance of ${E}$ yields ${y'(t_n) = 2^{-1/2} (n^2 - 1)}$ . By the mean value theorem, ${t_{n+1}-t_n \le \sqrt{2}(n^2-1)^{-1}}$ for ${n=2,3,\dots}$ . Since the series on the right converges, the limit ${T = \lim_{n\rightarrow\infty }t_n}$ is finite; this is the blow-up time.

But no blow-up occurs for the equation ${y''=-y^3}$ , where the nonlinear term pushes back toward equilibrium. Indeed, the invariant energy is now ${E= \frac12 (y')^2 + \frac14 y^4}$ , which implies that ${y}$ and ${y'}$ stay bounded. In the phase space ${(y,y')}$ the solution with initial values ${y(0)=1}$ , ${y'(0)=0}$ traces out this curve:

Closed trajectory in the phase space (y,y') — Closed trajectory in the phase space (y,y’)

Accordingly, the solution is a periodic function of ${t}$ (although this is not a trigonometric function):

Everything so far has been pretty straightforward. But here is a stranger animal: ${y'''=-y^3}$ . As in the previous example, nonlinear term pushes toward equilibrium. Using initial conditions ${y(0)=1}$ , ${y'(0)=y''(0)=0}$ , I get this numerical solution up to time ${T=5}$ :

As in the previous example, ${y}$ oscillates. But the speed and amplitude of oscillation are increasing.

Rapidly increasing:

In the plane ${(y,y')}$ the solution spirals out:

The plots make it clear that the solution ceases to exist in finite time, but I don’t have a proof. The issue is that the function ${y}$ does not escape to infinity in just one direction. Indeed, if ${y(t_0)>0}$ , then regardless of the values of ${y'}$ and ${y''}$ at ${t_0}$ , the negative third derivative will eventually make the function ${y}$ decrease, and so ${y}$ will attain the value ${0}$ at some ${t_1>t_0}$ . After that, the third derivative is positive, guaranteeing the existence of time ${t_2>t_1}$ when ${y}$ returns to ${0}$ again. I haven’t succeeded in proving that the limit ${\lim t_n}$ is finite, giving the time when oscillatory explosion occurs.

The plots were made in SageMathCloud:

var('t y y1 y2')
P = desolve_system_rk4([y1,y2,-y^3],[y,y1,y2], ics=[0,1,0,0],ivar=t,end_points=5,step=0.01)
Q=[ [i,j] for i,j,k,l in P]
list_plot(Q, plotjoined=True)

Oscillatory explosion

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