The equation I'm trying to find an expansion for is; $$ x^3 -(1+\epsilon)x^2 +\epsilon^2 +\epsilon^4 = 0$$

an asymptotic expansion for the perturbation equation using the ansatz $$x(\epsilon) = x_0 + x_1\epsilon + O(\epsilon ^2)$$ leads to the root $$x= 1+ \epsilon + O(\epsilon^2)$$

Now i'm trying to find the 2 other roots, using the usual method for a repeated root setting $$x = x_0 + \epsilon^{\alpha}X$$ with $$ x = O(1), \alpha>0$$ leads to $$\epsilon^{3\alpha}X^3 -(1+\epsilon)\epsilon^{2\alpha}X^2 +\epsilon^2 +\epsilon^4 = 0$$ and using the method of dominant balance leads to $$\epsilon^{2\alpha} = O(\epsilon^2) \implies \alpha = 1$$ and $$ \epsilon^3X^3-(1+\epsilon)\epsilon^2X^2 + \epsilon^2 + \epsilon^4 = 0$$

normally the shift would lead to a regular perturbation equation, but here it doesn't and using a perturbation expansion the scaled equation in the form $$X(\epsilon) = X_0+X_1\epsilon +O(\epsilon^2)$$ leads to results devoid of information (0=0), how should I go about answering questions of these types?

  • $\begingroup$ Neglecting terms in $\epsilon^2$ the roots are $1+\epsilon$ and $\pm\epsilon$ $\endgroup$
    – almagest
    Jan 21 '20 at 11:23

Your approach is working just fine.

At $O(1)$ you have

$$ x_0^3-x_0^2=-0$$

with solutions $x_0=0,0,1$. Then at $O(\epsilon)$ your equation depends on the value of $x_0$. If $x_0=1$, you get

$$ 3x_1-2x_1-1=0\Rightarrow x_1=1. $$

If $x_0=0$ then the next equation you get is at $O(\epsilon^2)$, and

$$ -x_1^2+1=0\Rightarrow x_1=\pm1. $$

So it depends a bit on what you want for your result, because $x_0=1+\epsilon$ satisfies the equation to within $O(\epsilon)$, but $x_0=\pm\epsilon$ satisfy the equation to within $O(\epsilon^2)$ (notice that $x=0$ satisfies the original equation to within $O(\epsilon^2)$), but all three approximations at within $O(\epsilon)$ of the real root.


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