I am dealing with the following non-linear differential equation:

$$\frac{d^2 x}{d t^2}=2\varepsilon\frac{d x}{d t}-\left(\frac{d x}{d t}\right)^3-x$$

I found that $x=0$ is the only one fixed point of that equation and it is:

  • stable for $\varepsilon\leq0$
  • unstable for $\varepsilon>0$

In particular, performing the stability analysis around the fixed point I used the perturbation ($a$ small):

$$x(t)=0+a\cdot e^{ibt}=a\cdot e^{\varepsilon t}e^{i\sqrt{1-\varepsilon^2 }t}$$

and so close to the bifucation point $\varepsilon\simeq0$ I have 2 characteristic timescales:


$\tau_2=\frac{1}{\sqrt{1-\varepsilon^2}}\simeq 1$

then, the text of the exercise says: "study the dynamics of the particle in the unstable regime ($\varepsilon > 0$), close to the bifurcation point $\varepsilon \ll 1$, using the multiple-scales method." First expand:

$$ x(t)=\varepsilon^{\frac{1}{2}}x_0(t)+\varepsilon^{\frac{3}{2}}x_1(t)+O(\varepsilon^{\frac{5}{2}}) $$ I cannot understand why the exercise text takes an expansion of that form (why does he need to use a fractional expansion)?

Then the exercise continue as follows: Using the analysis performed before, impose that the particle’s trajectory depends on two independent time scales: $t_1 = t$ and $t_2 = \varepsilon t$, so that $x_i(t_1 , t_2 )$.


Reading Bender & Orszag "Advanced Mathematical Methods for Scientists and Engineers" chapter 11, he says "The formal procedure consists of assuming a perturbation expansion of the form

$$y(t)=Y_0(t,\tau)+\varepsilon Y_1(t,\tau)+O(\varepsilon^2)$$

then he insert that expansion in the equation and collects powers of $\varepsilon$. This operation of collecting powers of $\varepsilon$ can be performed also using the expansion $$ x(t)=\varepsilon^{\frac{1}{2}}x_0(t)+\varepsilon^{\frac{3}{2}}x_1(t)+O(\varepsilon^{\frac{5}{2}}) $$

So it seems me that the powers of $\varepsilon$ in the expansion above does not matter (it sufficies that they differ in a way that allows them to be collected). Am I right?


1 Answer 1


It does matter. Consider $x=\epsilon^\alpha x_0+\epsilon^{1+\alpha}x_1+\ldots$, substituting into the equation gives (for the $x_0$ term, using dots for time derivatives), $$ \epsilon^\alpha\ddot x_0=2\epsilon^{1+\alpha}\dot x_0-\epsilon^{3\alpha}(\dot x_0)^3-\epsilon^\alpha x_0.$$ In order to solve this equation, you have to be able to balance at least 2 terms, and they have to be the 2 biggest terms.

Looking at the size of each term we get $\epsilon^\alpha$, $\epsilon^{1+\alpha}$, $\epsilon^{3\alpha}$, and $\epsilon^\alpha$.

There are 3 different choices of $\alpha$ that would balance the equation, $\alpha=1+\alpha$, $\alpha=3\alpha$, and $1+\alpha=3\alpha$. The first one is not possible, the second one gives no scaling, and the third choice gives $\alpha=1/2$.

If you check the differential equation, you will see that $\alpha=1/2$ makes the $2\epsilon^{1+\alpha}\dot x_0$ and $-\epsilon^{3\alpha}(\dot x_0)^3$ the two largest terms in the equation. The leading order equation would be $$2\dot x_0-(\dot x_0)^3=0.$$


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