# Asymptotic expansion with repeated roots

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?

• Neglecting terms in $\epsilon^2$ the roots are $1+\epsilon$ and $\pm\epsilon$ 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.