In a one dimensional setting, Hooke's Law, together with Newton's 2nd law, results in a differential equation of the form (setting the constants $m$ and $k$ equal to 1 for notational simplicity)

$x''(t) = -x(t)$,

with solutions of the form

$x(t) = a\sin(t) + b\cos(t)$,

where $a$ and $b$ are constants. The kinetic energy of this motion is bounded above by

$\frac{1}{2}(|a| + |b|)^2$.

But now consider the above differential equation with the sign reversed, namely

$x''(t) = x(t)$.

The general solution to this equation has the form

$x(t) = ae^t + be^{-t}$.

This motion seems strange to me, in part because the kinetic energy (and more simply, the speed) increases without bound as $t$ increases. Energy conservation is not violated, because the potential energy is negative, and decreases as the kinetic energy increases. Using two particles one can also set up a similar system of equations in which momentum is conserved as well. But this still seems odd, perhaps just because I haven't encountered such a phenomenon in everyday life.

Is this last equation a completely valid solution to Newton's 3 laws? Or is there some additional assumption I have overlooked in those laws that prevents these kinds of solutions? If not, is there some other principle that would rule out such solutions?

Finally, if this is a legitimate equation of motion given Newton's laws, is there a simple argument as to why we do not see such phenomena in everyday experience (in contrast to the behavior captured by Hooke's Law)?

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    $\begingroup$ I don't understand your question. You could always contrive some sort of second order differential equation that isn't found "in every day life". Does this mean the math is invalid? Math has never cared about whether or not it describes something physically realizable. Physicists put the constraints on the math depending on the system in question in order to describe reality. $\endgroup$ – BioPhysicist Mar 5 '19 at 8:22

This equation is not at all problematic. Kinetic energy will go to infinity only if you allow x to go to infinity, which you usually won't consider. Notice that this also occurs in the case of constant acceleration x''=g (like free fall on earth surface), or any potential decreasing in a strict monotonic way.

More intriguing is the case where the potential energy goes to minus infinity on a finite range, as is the case for the Coulombian interaction (V=-K/r) like in an atom. But even in this case the infinite energy comes from the imperfection of the model: inside the atomic nucleus, the potential is no longer Coulombian, but similar to a parabola with positive curvature.

**Edit occurence of equation x''=x. ** This equation is most often introduced as a model for an unstable equilibrium. Exactly in the same way as an oscillator, for small amplitude, can be modeled by an harmonic oscillator, by replacing the genuine potential by a parabola of positive curvature (with a spring strength given by the second derivative of the potential evaluated at its minimum).

A potential having a (local) maximum, can be modeled next to this equilibrium point by a parabola with negative curvature. A mobile at rest exactly at this point can stay at this point, but any minute disturbance will give to it an offset or a kick, thus inducing an increasing force in the same direction as the offset. So, instead of the recall to equilibrium appearing for an oscillator, one has an expelh force which in this model gives rise to an exponential escape. Of course , in such cases, the range of validity of this approximation is limited and x will likely not be able to go to infinity. The escape time is then given by \sqrt{m/V''} where V'' is the second derivative of the potential.

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  • $\begingroup$ Thanks, your comment on unstable equilibria largely answers my last question. You say "only if you allow x to go to infinity", but this is what I want to consider. The Coulomb potential is somewhat opposite to what I describe--it has a singularity at the origin, while the above potential has a singularity only at infinity. This allows unbounded growth of kinetic energy, without anything undefined. I was hoping there was some general principle that ruled out such behavior (for fundamental potentials, not models), e.g. a symmetry principle. Maybe the principle is just that V->0 as x->infinity? $\endgroup$ – Cooler Paradox Apr 26 '19 at 6:10

The equation you post corresponds to a Hamiltonian:

$$H = \frac{1}{2} (p^2 -x^2) $$

(the usual harmonic oscillator has a plus sign). Since the system is Hamiltonian it can't violate any of Newton's laws. However the Hamiltonian is unbounded below, which leads to unphysical infinite kinetic energies and as such it's unlikely that such a Hamiltonian is a good model for the system at large values of $x$. A classic example of a physical system is an inverted pendulum, whose Hamiltonian:

$$H = \frac{1}{2}p^2 - \cos(\theta) \approx \frac{1}{2}(p^2-x^2)$$

where the approximation is taking $x=(\theta-\pi)\ll 1$. Clearly as the exponential behaviour takes over $x$ becomes large and this approximation becomes poor, this is likely to be typical of any system experiencing such exponential growth.

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. . . just because I haven't encountered such a phenomenon in everyday life . . .

In addition to the existing answers, a common example of the form $x'' + \alpha x=0$ with $\alpha<0$ in physics is flutter in Fluid Structure Interaction. See here for a video of the phenomenon.

Of course, the governing equation is no longer valid when the system becomes "too nonlinear".

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$$x''(t) = x(t)\Rightarrow m\,x''(t) = m\,x(t)= \rm Force$$ is Newton's second law.

You have an ever increasing force acting on a mass $m$ so it is not surprising that the kinetic energy keeps increasing with time?

. . . . just because I haven't encountered such a phenomenon in everyday life . . . .

How such a situation is achieved in practice is another matter.
You need something which delivers an ever increasing amount of power as time goes on.

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