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):

Periodic solution, globally defined
Periodic solution, globally defined

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 plane (y,y')
The plane (y,y’)

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)

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