# Singular Points of DE

1. Nov 2, 2013

### FeDeX_LaTeX

The problem statement, all variables and given/known data
The equation of motion of a particle moving in a straight line is

$x'' - x + 2x^3 = 0$

and $x = \frac{1}{\sqrt{2}}, x' = u > 0$ at $t = 0$. Identify the singular points in the phase plane and sketch the phase trajectories. Describe the possible motions of the particle, indicating the ranges of u for which these motions occur.

The attempt at a solution

I really can't seem to get started on this question -- where are the singular points here, and how might I identify them? The co-efficient of x'' is 1, so it doesn't seem like I can divide by anything useful. I can't set a first derivative equal to zero, because there aren't any in this equation. Is there a form in which I have to rewrite this to get something useful out of it?

2. Nov 2, 2013

### tiny-tim

Hi FeDeX_LaTeX!
Multiply throughout by x' and then integrate?

3. Nov 2, 2013

### haruspex

Not an area I've previously worked in, but my reading of http://www.ucl.ac.uk/~ucahdrb/MATH3401/phaseplane.pdf is that the phase plane for a second order ODE is the plane (x', x). The 2nd order ODE is written as a first order ODE in terms of x and y where y = x'. Singular points refer to that context.

4. Nov 6, 2013

### FeDeX_LaTeX

Okay, with tiny-tim's suggestion, I had:

$$\frac{dx}{dt} \frac{d^{2}x}{dt^{2}} = (x - 2x^3) \frac{dx}{dt}$$

$$\implies \frac{1}{2} \frac{d}{dt} \left(\frac{dx}{dt}\right)^2 = (x - 2x^3) \frac{dx}{dt}$$

$$\implies \left(\frac{dx}{dt}\right)^2 = x^2 - x^4 + u^2 - \frac{1}{4}$$

after plugging in our initial conditions.

We must have $\left(\frac{dx}{dt}\right)^2 \geq 0$, so we have singularities whenever $x^2 - x^4 + u^2 - \frac{1}{4} < 0$, i.e. whenever $(x^2 - \frac{1}{2})^2 > u^2$. That'd give me something of the form x' = f(x) for which I can map a phase-plane plot -- all that remains is to re-arrange and integrate to get an equation in t and x, from which I could probably find the possible motions of the particle, correct?

5. Nov 6, 2013

### HallsofIvy

Staff Emeritus
This is just a matter of applying definitions. First, because this asks about "singular poinnts in the phase plane", we have to convert to the "phase plane"! Let v= x' so that x''= v' and the equation becomes $v'- x+ 2x^3= 0$ or $v'= x- 2x^3$.

So in the "phase plane" we have the pair of equatons $x'= v$, $v'= x- 2x^3$. A "singular point[/b] is where both x' and v' are 0. That is, where v= 0 and $x- 2x^3= 0$.