Rescaled Coordinates in a polynomial equation.

[Alone]

I have this question from Murdock's textbook called: "Perturbations: Methods and Theory":
Use rescaling to solve: $\phi(x,\epsilon) = \epsilon x^2 + x+1 = 0$ and $\varphi (x,\epsilon) = \epsilon x^3+ x^2 - 4=0$.

I'll write my attempt at solving these two equations, first the first polynomial.

According to the book I first need to find an undetermined gauge that solves the above equation when $\epsilon = 0$, which means I need to guess $x \approx -1 +y \delta_1(\epsilon)$, now I plug it back to the above equation to find the following equation:
$$\epsilon -2y\epsilon \delta_1 + y^2 \delta_1 \epsilon +y \delta = 0 $$

Now my problem is in finding a suitable gauge for $\delta_1(\epsilon)$ that solves the above last equation, it should be proportional to some power of $\epsilon$, i.e $\epsilon^r$, such that higher powers of $\epsilon$ in the last equation get neglected and we can find a solution for $y$ which gives a non-zero solution to $y$, but I don't see it.

P.S
@Carla1985 you might know how to approach this question if I remember correctly you posted questions on perturbation method and asymptotics.

Cheers!
https://mathhelpboards.com/members/carla1985/

 

[jvicens]

Thank you for your question. The rescaling method is a useful tool in solving perturbation problems, but it can be tricky to find the appropriate gauge function. Let me walk you through the steps for solving these two equations using rescaling.

For the first equation, let's start by setting $\delta_1(\epsilon) = \epsilon^r$ and plug it into the equation. This gives us:
$$\epsilon - 2y\epsilon^{r+1} + y^2 \epsilon^{2r+1} + y\epsilon^r = 0$$
Now, in order for the higher powers of $\epsilon$ to be negligible, we need to have $r+1 = 0$ and $2r+1 = 0$. Solving these equations gives us $r = -1$ and $y = 1$. Therefore, our gauge function is $\delta_1(\epsilon) = \epsilon^{-1}$.

Substituting this into our original equation, we get:
$$\epsilon - 2\epsilon^{-1} + \epsilon^{-2} + \epsilon^{-1} = 0$$
Multiplying both sides by $\epsilon^2$, we get:
$$\epsilon^3 - 2\epsilon + 1 + \epsilon^2 = 0$$
This is a cubic equation, which can be solved using standard methods. The solutions are $x = -1, -1 \pm \sqrt{2}$. Therefore, the solutions to the original equation are:
$$x = -1 + \epsilon^{-1}, -1 \pm \sqrt{2} + \epsilon^{-1}$$

Now, let's move on to the second equation. Again, we start by setting $\delta_1(\epsilon) = \epsilon^r$ and plug it into the equation. This gives us:
$$\epsilon - 4y\epsilon^{r+1} + y^2 \epsilon^{2r+1} + y^3 \epsilon^{3r+1} + y^2 \epsilon^{2r+1} - 4 = 0$$
To make the higher powers of $\epsilon$ negligible, we need to have $r+1 = 0$, $2r+1 = 0$ and $3r+1 = 0$. Solving these equations gives us $r = -1