MHB Solve Diff Eq: Change of Coordinates to Eliminate Squared Terms

[Fantini]

Here is the question:

Consider the differential equation

$$x' = a_1 x + a_2 x^2 + a_3 x^3 + \cdots,$$

with $a_1 \neq 0$. Show that there exists a $C^2$ change of coordinates of the form $x = y + \alpha y^2$ that rewrites the equation (locally around $x=0$) as

$$y' = a_1 y + b_3 y^3 + \cdots,$$

that is, that eliminates the squared term.

I have no idea how to go about it.

 

[I like Serena]

Here is the question:

Consider the differential equation

$$x' = a_1 x + a_2 x^2 + a_3 x^3 + \cdots,$$

with $a_1 \neq 0$. Show that there exists a $C^2$ change of coordinates of the form $x = y + \alpha y^2$ that rewrites the equation (locally around $x=0$) as

$$y' = a_1 y + b_3 y^3 + \cdots,$$

that is, that eliminates the squared term.

I have no idea how to go about it.

Substitute $x = y + \alpha y^2$ into the differential equation?

 

[TheBigBadBen]

Here is the question:

Consider the differential equation

$$x' = a_1 x + a_2 x^2 + a_3 x^3 + \cdots,$$

with $a_1 \neq 0$. Show that there exists a $C^2$ change of coordinates of the form $x = y + \alpha y^2$ that rewrites the equation (locally around $x=0$) as

$$y' = a_1 y + b_3 y^3 + \cdots,$$

that is, that eliminates the squared term.

I have no idea how to go about it.

Substitute $x = y + \alpha y^2$ into the differential equation?

First of all, for convenience, I will write $p(x) \equiv a_1 x + a_2 x^2 + a_3 x^3 + \cdots $

Now, if you try substituting in, you get
$$
(y + \alpha y^2)' = p(y + \alpha y^2)
$$
First of all, note that $(y + \alpha y^2)' = (2\alpha y + 1)y' $. As for the other side, we can note that
$$
p(y + \alpha y^2) =
a_1(y + \alpha y^2) +
a_2(y^2 + 2\alpha y^3 + \alpha ^2 y^4) + \cdots
$$

I'm not sure where to go from there either, but I think it might help to have it down. I suppose it might have something to do with finding a condition on $\alpha$ so that $(2\alpha y + 1)$ divides $p(y + \alpha y^2)$

 

[Opalg]

First of all, for convenience, I will write $p(x) \equiv a_1 x + a_2 x^2 + a_3 x^3 + \cdots $

Now, if you try substituting in, you get
$$
(y + \alpha y^2)' = p(y + \alpha y^2)
$$
First of all, note that $(y + \alpha y^2)' = (2\alpha y + 1)y' $. As for the other side, we can note that
$$
p(y + \alpha y^2) =
a_1(y + \alpha y^2) +
a_2(y^2 + 2\alpha y^3 + \alpha ^2 y^4) + \cdots
$$

Following TheBigBadBen's approach, $(2\alpha y + 1)y' = a_1(y + \alpha y^2) + a_2(y^2 + 2\alpha y^3 + \alpha ^2 y^4) + \ldots$. Now use the binomial expansion of $(1 + 2\alpha y)^{-1}$ (assuming that it converges) to get $$\begin{aligned}y' &= \bigl( a_1(y + \alpha y^2) + a_2(y^2 + 2\alpha y^3 + \alpha ^2 y^4) + \ldots\bigr)(1 + 2\alpha y)^{-1} \\ &= \bigl( a_1(y + \alpha y^2) + a_2(y^2 + 2\alpha y^3 + \alpha ^2 y^4) + \ldots\bigr)(1 - 2\alpha y + 4\alpha^2y^2 - \ldots) \\ &= a_1y + (a_2 - a_1\alpha)y^2 + \ldots.\end{aligned}$$ You can now put the coefficient of $y^2$ equal to $0$ to get an equation for $\alpha$. But that is a rather heuristic, non-rigorous method. I'm not sure how to deal with the requirement that this should be a $C^2$ change of coordinates.

 

[Fantini]

This is what I thought as well, but he said we would use something like the inverse function theorem or the implicit function theorem. How would we apply any of these there?

 

[I like Serena]

But that is a rather heuristic, non-rigorous method. I'm not sure how to deal with the requirement that this should be a $C^2$ change of coordinates.

I believe that $x=y+\alpha y^2$ is a $C^\infty$ change of coordinates.
Furthermore, if both series are convergent within some area around zero, they will be the same.

This is what I thought as well, but he said we would use something like the inverse function theorem or the implicit function theorem. How would we apply any of these there?

I don't see how you might use either of those theorems in a useful manner.

 

[Fantini]

I believe this settles the question. I don't think there's much more to be done than what we did so far. Thanks! :)