Consider the mass-spring system governed by the differential equation, $$ m\ddot{x}=-F(x) $$

Where $x(t)$ is the time-dependent position displacement of the mass. The spring force is a nonlinear function of $x$ and given by, $$F(x)=\alpha x-\beta x^3= \frac{dV}{dx}$$ Where $\alpha$ and $\beta$ are positive constants.

Sketch the function $V(x)$ and show from this plot that the first equation has three stationary points. Can you judge from the plot of $V(x)$ which of these stationary points are stable and which are unstable? Assume that $V(0)=0$

Now by integrating $\frac{dV}{dx}$, I got $$V(x)=\frac{1}{2} \alpha x^2-\frac{1}{4}\beta x^4 $$

We see that $V(x)$ is the potential energy of the mass as a function of displacement. Plotting this gave me something like this

enter image description here

Where stationary points are at $x=0, x=\pm \sqrt{\frac{2\alpha}{\beta}}$. I got stuck here and according to the provided solution, the local minimality of $V(x)$ at $0$ implies stability and the maximality of $V(x)$ at $x=\pm \sqrt{\frac{2\alpha}{\beta}}$ implies instability. There is no further explanation and I am extremely confused. I am also not sure how showing $V(x)$ is stationary at the given points shows that $x(t)$ is also stationary.

I understand that if $\frac{dV}{dx}=0$ then $\ddot{x}=0$. But my understanding was that $\dot{x}$ also needs to be $0$ for $x(t)$ to be stationary. In fact for $V(x)$ where $x=0$ which is the local minima, $\dot{x}$ cannot be $0$ since If the potential energy is at its minimum, the kinetic energy has to be at its maximum value (conservation of energy) which means that $\dot{x}$ has to be the maximum value. So $\frac{dx}{dt}\neq 0$ which contradicts with my understanding that $\frac{dx}{dt}$ must equal $0$ at stationary points.

I would appreciate any help clarifying my confusion.

Edit: Three dimensional phase portrait with $x$, $\dot{x}$ and $\ddot{x}$ as axis.

enter image description here

  • 1
    $\begingroup$ Think of the graph as a height profile, a landscape, and let a ball roll from any position in question. It stays at the minimum and any slight perturbation at a maximum will make it roll off. $\endgroup$ – LutzL Oct 27 '18 at 19:49


$$ m\dot x\ddot x + F(x)\dot x = 0\to \frac 12\frac{d}{dt}(m\dot x)^2+\frac{d}{dt}\left(\frac 12 \alpha x^2-\frac 14 \beta x^4\right) = 0 $$

with the integral

$$ m\dot x^2+\alpha x^2-\frac 12 \beta x^4 = C_0 $$

Equilibrium is attained when $\dot x = 0$ or

$$ \alpha x^2 - \frac 12 \beta x^4= C_0 $$

or at

$$ x^* = \pm \sqrt{\frac{\alpha \pm\sqrt{\alpha ^2-2 \beta C_0}}{\beta }} $$

The equilibrium qualification can be done depending on $\alpha^2-2\beta C_0$ and also on characterizing those feasible points as sources saddle points or sinks.


If $C_0 = 0$ then

$$ x^* = \left\{-\sqrt{\frac{2\alpha}{\beta}},0,\sqrt{\frac{2\alpha}{\beta}}\right\} $$

Attached a plot for $m = 1,\alpha = 1, \beta = 1$ showing the phase plane movement partial orbits, with $C_0 = 1$ (red), $C_0 = 0.5$ (blue) and $C_0 = 0.2$ (green)

enter image description here

Including viscous dissipation as $-k \dot x$ with $k = \frac 18$ we have the path from $\dot x(0) = 0.7, x(0) = 0$ (in red)

enter image description here


Near $x = 0$ the linearized dynamic system behaves as

$$ m\ddot x + \alpha x = 0 $$

with solution

$$ x_0(t) = C_1 \sin \left(\frac{\sqrt{\alpha } t}{\sqrt{m}}\right)+C_2 \cos \left(\frac{\sqrt{\alpha } t}{\sqrt{m}}\right) $$

and near $x = \pm\sqrt{\frac{2\alpha}{\beta}}$ the linearized movement is

$$ m\ddot x -2\alpha x = 0 $$

with solution

$$ x_{\sqrt{\frac{2\alpha}{\beta}}}(t) = C_1 e^{\frac{\sqrt{2\alpha } t}{\sqrt{m}}}+C_2 e^{-\frac{\sqrt{2\alpha } t}{\sqrt{m}}}\ \ \mbox{(unstable)} $$

  • $\begingroup$ Thanks for your answer. in this case, what would the physical interpretation of $C_{0}$ be? It seems to me that it represents the total energy of the system. But in that case, how can we conclude that $C_{0}=0$? $\endgroup$ – Sei Sakata Oct 27 '18 at 20:39
  • 1
    $\begingroup$ @SeiSakata Yes. The $C_0$ is the total energy which remains constant as far as this system is a conservative one. We have $\frac 12 m \dot x^2 + V(x) = C_0$ If $C_0 = 0$ this means that the point is located at one of it's equilibrium points. $\endgroup$ – Cesareo Oct 27 '18 at 20:46
  • $\begingroup$ So as long if $x=$ one of the solutions, and there is no external force, the mass will remain stationary and no movement whatsoever? Is this the definition of equilibrium state? It is extremely counterintuitive to me that there can be a such point other than the $x=0$ where the extension of the spring is zero. $\endgroup$ – Sei Sakata Oct 27 '18 at 22:26
  • $\begingroup$ @SeiSakata In a conservative system the movement can assume diverse paths as shown in the attached picture. Introducing dissipation, the energy will decrease with time, finishing in a stable equilibrium point. $\endgroup$ – Cesareo Oct 27 '18 at 23:50
  • $\begingroup$ Or you need a controller that effects minute corrections to stay in an instable stationary point, like balancing a broom stick upright on your finger. $\endgroup$ – LutzL Oct 28 '18 at 9:44

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