Welcome to our community

Be a part of something great, join today!

Change of coordinates

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
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.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
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

Active member
May 12, 2013
84
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

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
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

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
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?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
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

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
I believe this settles the question. I don't think there's much more to be done than what we did so far. Thanks! :)